Author Archives: Swati

Exercise 13.1

Question 1.
A student buys a pen for Rs. 90 and sells it for Rs. 100. Find his gain and gain percent ?
Solution:
C.P. of a pen = Rs. 90
and S.P. = Rs. 100
Gain = S.P. – C.P. = Rs. 100 – 90 (S.P > C.P.)
= Rs. 10

Question 2.
Rekha bought a saree for Rs. 1240 and sold it for Rs. 1147. Find her loss and loss percent.
Solution:
C.P. of saree.= Rs. 1240 and
S.P. = Rs. 1147
Loss = C.P – S.P. = Rs. 1240 – Rs. 1147 (C.P. > S.P.)
= Rs. 93

Question 3.
A boy buys 9 apples for Rs. 9.60 and sells them at 11 for Rs. 12. Find his gain or loss percent.
Solution:
L.C.M. of 9 and 11 = 99
Let 99 apples were purchased.
C.P. of 99 apples at the rate of 9 apples for Rs. 9.60

Question 4.
The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percent.
Solution:
C.P. of 10 articles = S.P. of 9 articles = 90 (Suppose)

Question 5.
A retailer buys a radio for Rs. 225. His overhead expenses are Rs. 15. If he sells the radio for Rs. 300, determine his profit percent.
Solution:
Cost of radio = Rs. 225
Over head expenses = Rs. 15
Total C.P. of the ratio = Rs. 225 + 15 = Rs. 240
S.P. of radio = Rs. 300
Gain = S.P. – C.P. = Rs. 300 – Rs. 240 = Rs. 60

Question 6.
A retailer buys a cooler for Rs. 1200 and overhead expenses on it are Rs. 40. If he sells the cooler for Rs. 1550, determine his profit percent.
Solution:
Cost of cooler = Rs. 1200
Overhead expenses = Rs. 40
Total cost price of cooler = Rs. 1200 + Rs. 40 = Rs. 1240
Selling price = Rs. 1550
Gain = S.P. – C.P. = Rs. 1550 – Rs. 1240 = Rs. 310

Question 7.
A dealer buys a wristwatch for Rs. 225 and spends Rs. 15 on its repairs. If he sells the same for Rs. 300, find his profit percent.
Solution:
Cost of wristwatch = Rs. 225
Cost on repairs = Rs. 15
Total cost price = Rs. 225 + Rs. 15 = Rs. 240
Selling price = Rs. 300
Gain = S.P. – C.P. = Rs. 300 – Rs. 240 = Rs. 60

Question 8.
Ramesh bought two boxes for Rs. 1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.
Solution:
Total cost price of two boxes = Rs. 1300
S.P. of each box is same.
Let S.P of each box = Rs. 100
S.P. of first box = Rs. 100
Gain = 20%

Question 9.
If the selling price of 10 pens is equal to cost price of 14 pens, find the gain percent.
Solution:
S.P. of 10 pens = C.P. of 14 pens = Rs. 100 (Suppose)

Question 10.
If the cost price of 18 chairs be equal to selling price of 16 chairs, And the gain or loss percent.
Solution:
C.P. of 18 chairs = S.P. of 16 chairs = Rs. 100 (Suppose)

Question 11.
If the selling price of 18 oranges is equal to the cost price of 16 oranges, And the loss percent.
Solution:
S.P. of 18 oranges = C.P. of 16 oranges = Rs. 100 (Suppose)

Question 12.
Ravish sold his motorcycle to Vineet at a loss of 28%. Vineet spent Rs. 1680 on its repairs and sold the motor cycle to Rahul for Rs. 35910, thereby making a profit of 12.5%, find the cost price of the motor cycle for Ravish.
Solution:
Cost price of cycle for Rahul or Selling price for Vineet = Rs. 35910
Gain = 12.5%

Question 13.
By selling a book for Rs. 258, a book-seller gains 20%. For how much should he sell it to gain 30% ?
Solution:
S.P. of a book = Rs. 258
Gain = 20%

Question 14.
A defective briefcase costing Rs. 800 is being sold at a loss of 8%. If its price is further reduced by 5%, find its selling price.
Solution:
C.P. of a briefcase = Rs. 800
Loss = 8%

Question 15.
By selling 90 ball pens for Rs. 160, a person losses 20%. How many ball pens should be sold for Rs. 96 so as to have a profit of 20%.
Solution:
S.P of 90 ball pens = Rs. 160

Question 16.
A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs. 36.75 less, he would have gained 30%. Find the cost price of the article.
Solution:
Let C.P. of the article = Rs. 100
In first case gain = 25%
S.P. = 100 + 25 = Rs. 125
In second case,
C.P. = 20% less of Rs. 100 = 100 – 20 = Rs. 80
Gain = 30%

Question 17.
A dishonest shopkeeper professes to sell pulses at his cost price but uses a false weight of 950 for each kilogram. Find his gain percent.
Solution:
Let C.P. of 1 kg of pulses = Rs. 100
S.P. of 950 gm = Rs. 100

Question 18.
A dealer bought two tables for Rs. 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then he found that each table was sold for the same price, find the cost price of each table.
Solution:
Cost price of two tables = Rs. 3120
Let S.P of each table = Rs. 100
Now S.P of one table = Rs. 100
Loss = 15%

Question 19.
Mariam bought two fans for Rs. 3605. She sold one at a profit of 15% and the other at a loss of 9%. If Mariam obtained the same amount for each fan, find the cost price of each fan.
Solution:
Total cost price of two fans = Rs. 3605
Let selling price of each fan = Rs. 100
Now S.P. of first fan = Rs. 100
Profit = 15%

Question 20.
Some toffees are bought at the rate of 11 for Rs. 10 and the same number at the rate of 9 for Rs. 10. If the whole lot is sold at one rupee per toffee, find the gain or loss percent on the whole transaction.
Solution:
L.C.M. of 11 and 9 = 99
Let each time 99 toffees are bought.
In first case, the C.P. of 99 toffees at the rate of 11 for Rs. 10

Question 21.
A tricycle is sold at a gain of 16%. Had it been sold for Rs. 100 more, the gain would have been 20%. Find the C.P. of the tricycle.
Solution:
Let the C.P. of tricycle = Rs. 100
In first case, gain = 16%
S.P. = Rs. 100 + 16 = Rs. 116
In second case, gain = 20%
S.P. = Rs. 100 + 20 = Rs. 120
Difference in S.P.’s = Rs. 120 – Rs. 116 = Rs. 4
If difference is Rs. 4,
then C.P. of the tricycle = Rs. 100
and if difference is Re. 1, then C.P.

Question 22.
Shabana bought 16 dozen ball pens and sold then at a loss equal to S.P. of 8 ball pens. Find
(i) her loss percent
(ii) S.P. of 1 dozen ball pens, if she purchased these 16 dozen ball pens for Rs. 576.
Solution:
C.P. of 16 dozen ball pens = S.P. of 16 dozen pens – loss
C.P. of 16 x 12 pens = S.P. of 16 dozen pens x S.P. of 8 pens
C.P. of 192 pens = S.P. of 16 x 12 pens x S.P. of 8 pens
S.P. of 192 pens + S.P. of 8 pens = S.P. of 200 pens
(i) Let C.P. of 1 pen = Re. 1
Then C.P. of 192 pens = Rs. 192
and S.P. of 200 pens = Rs. 192

Question 23.
The difference between two selling prices of a shirt at profits of 4% and 5% is Rs. 6. Find
(i) C.P. of the shirt
(ii) The two selling prices of the shirt
Solution:
The difference of two selling prices of shirt = Rs. 6
Difference in profits of 4% and 5% = 5 – 4 = 1%
(i) C.P. = 1 x 6 x 100 = Rs. 600

Question 24.
Toshiba bought 100 hens for Rs. 8000 and sold 20 of these at a gain of 5%. At what gain percent she must sell the remaining hens so as to gain 20% on the whole ?
Solution:
Total number of hens bought = 100
C.P. of 100 hens = Rs. 8000

Exercise 13.2

Question 1.
Find the S.P. If
(i) M.P. = Rs. 1300 and Discount = 10%
(ii) M.P. = Rs. 500 and Discount = 15%
Solution:
(i) M.P. = Rs. 1300, Discount = 10%

Question 2.
Find the M.P. If
(i) S.P. = Rs. 1222 and Discount = 6%
(ii) S.P. = Rs. 495 and Discount = 1%
Solution:
(i) S.P. = Rs. 1222, discount = 6%

Question 3.
Find discount in percent when
(i) M.P. = Rs. 900 and S.P. = Rs. 873
(ii) M.P. = Rs. 500 and S.P. = Rs. 425
Solution:
(i) M.P. = Rs. 900
S.P. = Rs. 873
Discount = M.P. – S.P. = Rs. 900 – Rs. 873 = Rs. 27

(ii) M.P. = Rs. 500
S.P. = Rs. 425
Discount M.P. – S.P. = Rs. 500 – Rs. 425 = = Rs. 75

Question 4.
A shop selling sewing machines offers 3% discount on ail cash purchases. What cash amount does a customer pay for a sewing machine, the price of which is marked as Rs. 650.
Solution:
Marked price (M.P.) of one sewing machine = Rs. 650
Rate of discount = 3%

Question 5.
The marked price of a ceiling fan is Rs. 720. During off season, it is sold for Rs. 684. Determine the discount percent.
Solution:
Marked price (M.P.) of fan = Rs. 720
Sale price (S.P.) = Rs. 684
Amount of discount = M.P. – S.P. = Rs. 720 – Rs. 684 = Rs. 36

Question 6.
On the eve of Gandhi Jayanti, a saree is sold for Rs. 720 after allowing 20% discount. What is the marked price ?
Solution:
S.P. of saree = Rs. 720
Rate of discount = 20%

Question 7.
After allowing a discount of 712 % on the marked price, an article is sold for Rs. 555. Find its marked price.
Solution:
Selling price (S.P.) = Rs. 555

Question 8.
A shopkeeper allows his customers 10% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs. 250?
Solution:
Marked price = Rs. 250
Discount allowed = 10%

Question 9.
A shopkeeper allows 20% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs. 500 ?
Solution:
Marked price (M.P.) of an article = Rs. 500
Discount allowed = 20%
Selling price (S.P.)

Question 10.
A tradesman marks his goods at such a price that after allowing a discount of 15%, he makes a profit of 20%. What is the marked price of an article whose cost price is Rs 170 ?
Solution:
Rate of discount = 15% gain = 20%
Cost price (C.P.) of an article = Rs 170
Selling price (S.P.)

Question 11.
A shopkeeper marks his goods in such a way that after allowing a discount of 25% on the marked price, he still makes a profit of 50%. Find the ratio of the C.P. to the M.P.
Solution:
Let cost price (C.P.) = Rs 100
Profit = 50%
Selling Price (S.P.)

Question 12.
A cycle dealer offers a discount of 10% and still makes a profit of 26%. What is the actual cost to him of a cycle whose marked price is Rs 840 ?
Solution:
Rate of discount = 10%
Gain = 26%
Marked Price (M.P.) = Rs 840
Selling Price (S.P.)

Question 13.
A shopkeeper allows 23% commission on his advertised price and still makes a profit of 10%. If he gains Rs 56 on one item, find his advertised price.
Solution:
Rate of commission = 23%
Profit = 10%
Total gain = Rs 56

Question 14.
A shopkeeper marks his goods at 40% above the cost price but allows a discount of 5% for cash payment to his customers. What actual profit does he make, if he receives Rs 1064 after paying the discount ?
Solution:
Let cost price (C.P.) = Rs 100
Marked price = Rs 100 + 40 = Rs 140
Rate of discount = 5%
Selling price (S.P)

Question 15.
By selling a pair of ear rings at a discount of 25% on the marked price, a jeweller makes a profit of 16%. If the profit is Rs 48, what is the cost price ? What is the marked price and the price at which the pair was eventually bought ?
Solution:
Total profit = Rs 48
Profit percent = 16%
Cost price = 48×10016 = Rs 300
Selling Price = C.P. + profit = Rs 300 + Rs 48 = Rs 348
Rate of discount = 25%

Question 16.
A publisher gives 32% discount on the printed price of a book to booksellers. What does a bookseller pay for a book whose printed price is Rs 275 ?
Solution:
Printed price of a book = Rs 275
Rate of discount = 32%
Selling price (S.P.)

Question 17.
After allowing a discount of 20% on the marked price of a lamp, a trader loses 10%. By what percentage is the marked price above the cost price ?
Solution:
Rate of discount = 20%
Loss = 10%
Let the cost price of the lamp = Rs 100
Loss = 10%
Selling price = Rs 100 – 10 = Rs 90
Rate of discount = 20%

Question 18.
The list price of a table fan is Rs 480 and it is available to a retailer at 25% discount. For how much should a retailer sell it to gain 15% ?
Solution:
List price of table fan (M.P.) = Rs 480
Rate of discount = 25%
Selling price (S.P)

Question 19.
Rohit buys an item at 25% discount on the marked price. He sells it for Rs 660, making a profit of 10%. What is the marked price of the item ?
Solution:
Rate of discount = 25%
Selling price (S.P.) for Rohit = Rs 660
Profit = 10%

Question 20.
A cycle merchant allows 20% discount on the marked price of the cycles and still makes a profit of 20%. If he gains Rs 360 over the sale of one cycle, find the marked price of the cycle.
Solution:
Rate of discount = 20%
Profit = 20%
Total gain = Rs 360

Question 21.
Jyoti and Meena run a ready-made garment shop. They mark the garments at such a price that even after allowing a discount of 12.5%, they make a profit of 10%. Find the marked price of a suit which costs them Rs 1470.
Solution:
Rate of discount = 12.5%
Profit = 10%
Cost price = Rs 1470

Question 22.
What price should Aslam mark on a pair of shoes ? Which costs him Rs 1200 so as to gain 12% after allowing a discount of 16% ?
Solution:
Cost price of shoes = Rs 1200
Gain = 12%

Question 23.
Jasmine allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a shirt for her marked at Rs 850 ?
Solution:
Marked price of a shirt = Rs 850
Discount = 4%
Selling price

Question 24.
A shopkeeper offers 10% off-season discount to the customers and still makes a profit of 26%. What is the cost price for the shopkeeper on a pair of shoes marked at Rs 1120 ?
Solution:
Marked price = Rs 1120
Rate of discount = 10%
S.P. of shoes

Question 25.
A lady shopkeeper allows her customers 10% discount on the marked price of the goods and still gets a profit of 25%. What is the cost price of a fan for her marked at Rs 1250 ?
Solution:
Marked price (M.P.) of fan = Rs 1250
Discount = 10%
S.P. of the fan

Exercise 13.3

Question 1.
The list price of a refrigerator is Rs 9700. If a value added tax of 6% is to be charged on it, how much one has to pay to buy the refrigerator ?
Solution:
List price of refrigerator = Rs 9700
Rate of VAT = 6%
Amount of VAT = Rs. 9700×6100 = Rs 582
Total price to be paid = Rs 9700 + 582 = Rs 10282

Question 2.
Vikram bought a watch for Rs 825. If this amount includes 10% VAT on the list price, what was the list price of the watch ?
Solution:
Price of watch including VAT = Rs 825
Rate of VAT = 10%

Question 3.
Aman bought a shirt for Rs 374.50 which includes 7% VAT. Find the list price of the shirt.
Solution:
Cost price of the shirt = Rs 374.50
Rate of VAT = 7%

Question 4.
Rani purchases a pair of shoes whose sale price is Rs 175. If he pays a VAT at the rate of 7%, how much amount does he pays as VAT. Find the net value of the pair of shoes.
Solution:
Sale price of a pair of shoes = Rs 175
Rate of VAT = 7%
Total VAT paid = Rs. 175×7100 = Rs 12.25
and total amount paid = Rs 175 + Rs 12.25 = Rs 187.25https://googleads.g.doubleclick.net/pagead/ads?client=ca-pub-7601472013083661&output=html&h=280&adk=2523109437&adf=980553861&pi=t.aa~a.4089255474~i.16~rp.4&w=750&fwrn=4&fwrnh=100&lmt=1642576099&num_ads=1&rafmt=1&armr=3&sem=mc&pwprc=1894297687&psa=1&ad_type=text_image&format=750×280&url=https%3A%2F%2Fwww.learninsta.com%2Frd-sharma-class-8-solutions-chapter-13-profits-loss-discount-and-value-added-tax-vat-ex-13-3%2F&flash=0&fwr=0&pra=3&rh=188&rw=750&rpe=1&resp_fmts=3&wgl=1&fa=27&adsid=ChEIgN6ykAYQlqfqreX-0tatARI5AOa2GRQuFTRuP5gxgF63-8ooeukU8cZ7QCQZcMs0Bo08KBiUSCO_wUJ4DWFWZv_YD9YpnCl_xW2w&uach=WyJXaW5kb3dzIiwiMTAuMC4wIiwieDg2IiwiIiwiOTguMC40NzU4LjEwMiIsW10sbnVsbCxudWxsLCI2NCIsW1siIE5vdCBBO0JyYW5kIiwiOTkuMC4wLjAiXSxbIkNocm9taXVtIiwiOTguMC40NzU4LjEwMiJdLFsiR29vZ2xlIENocm9tZSIsIjk4LjAuNDc1OC4xMDIiXV1d&dt=1645025146248&bpp=2&bdt=6721&idt=2&shv=r20220214&mjsv=m202202090101&ptt=9&saldr=aa&abxe=1&cookie=ID%3D1be37a61a5e7cca6-2207ff21a8d0005c%3AT%3D1643370310%3ART%3D1645025060%3AS%3DALNI_MZlVwsQQNfk-mglMlxYK4Mv7pUwgg&prev_fmts=0x0%2C750x280%2C750x280&nras=3&correlator=3657959876739&frm=20&pv=1&ga_vid=528868594.1643370309&ga_sid=1645025145&ga_hid=193832627&ga_fc=1&u_tz=330&u_his=4&u_h=715&u_w=1270&u_ah=681&u_aw=1270&u_cd=24&u_sd=1.513&dmc=8&adx=57&ady=2221&biw=1254&bih=610&scr_x=0&scr_y=667&eid=42531397%2C44750773%2C21067496&oid=2&pvsid=1097847813229336&pem=972&tmod=1825076183&uas=3&nvt=1&ref=https%3A%2F%2Fwww.learninsta.com%2Frd-sharma-class-8-solutions%2F&eae=0&fc=1408&brdim=0%2C0%2C0%2C0%2C1270%2C0%2C1270%2C681%2C1270%2C610&vis=1&rsz=%7C%7Cs%7C&abl=NS&fu=128&bc=31&jar=2022-02-16-15&ifi=4&uci=a!4&btvi=3&fsb=1&xpc=BHm11P10xT&p=https%3A//www.learninsta.com&dtd=65

Question 5.
Swarna paid Rs 20 as VAT on a pair of shoes worth Rs 250. Find the rate of VAT.
Solution:
Amount of VAT paid = Rs 20
Price of the pair of shoes = Rs 250
Rate of VAT = 20×100250 = 8%

Question 6.
Sarita buys goods worth Rs 5,500. She gets a rebate of 5% on it. After getting the rebate if VAT at the rate of 5% is charged, find the amount she will have to pay for the goods.
Solution:
Price of goods = Rs 5,500
Rate of rebate = 5%
Sales price after rebate

= Rs 5,225
Rate of VAT = 5%
Amount of VAT

Amount to be paid after paying VAT = Rs 5,225 + 261.25 = Rs 5486.25

Question 7.
The cost of furniture inclusive of VAT is Rs 7,150. If the rate of VAT is 10%, find the original cost of the furniture.
Solution:
Cost of furniture including VAT = Rs 7,150
Rate of VAT = 10%
Original cost of furniture

Question 8.
A refrigerator is available for Rs 13,750 including VAT. If the rate of VAT is 10%, find the original cost of refrigerator.
Solution:
Cost of refrigerator including VAT = Rs 13,750
Rate of VAT = 10%
Actual price of refrigerator

Question 9.
A colour TV is available for Rs 13,440 inclusive of VAT. If the original cost of TV is Rs 12,000, find the rate of VAT.
Solution:
Cost of TV including VAT = Rs 13,440
Actual cost = Rs 12,000
Amount of VAT = Rs 13,440 – Rs 12,000 = Rs 1,440

Question 10.
Reena goes to a shop to buy a radio, costing Rs 2,568. The rate of VAT is 7%. She tells the shopkeeper to reduce the price of the radio such that she has to pay Rs 2,568 inclusive of VAT. Find the reduction needed in the price of radio.
Solution:
Price of radio inclusive of VAT = Rs 2,568
Rate of VAT = 7%

But the cost of radio in the beginning = Rs 2,568
Reduction = Rs 2568 – Rs 2,400 = Rs 168

Question 11.
Rajat goes to a departmental store and buys the following articles :

Calculate the total amount he has to pay to the store.
Solution:
Price of 2 pairs of shoes @ Rs 800 = Rs800 x 2 = Rs 1,600
Rate of VAT = 5%


Total amount to be paid = Rs 1,680 + Rs 1,590 + Rs 1,352 = Rs 4,622

Question 12.
Ajit buys a motorcycle for Rs 17,600 including value added tax. If the rate of VAT is 10%, what is the sale price of the motorcycle ?
Solution:
Cost price of motorcycle (including VAT) = Rs 17,600
Rate of VAT = 10%
Sale price of motorcycle

Question 13.
Manoj buys a leather coat costing Rs 900 at Rs 990 after paying the VAT. Calculate the VAT charged on the cost.
Solution:
Cost price of coat = Rs 900
And sale price including VAT = Rs 990
Amount of VAT = Rs 990 – Rs 900 = Rs 90
Rate of VAT = 90×100900 = 10%

Question 14.
Rakesh goes to a departmental store and purchases the following articles:
(i) biscuits and bakery products costing Rs 50, VAT @ 5%.
(ii) medicines costing Rs 90. VAT @ 10%,
(iii) clothes costing Rs 400, VAT @ 1% and
(iv) cosmetics costing Rs 150, VAT @ 10% Calculate the total amount to be paid by Rakesh to the store.
Solution:
(i) Cost of biscuits and bakery product = Rs 50
VAT = 5%

Total amount of the bill = Rs 52.50 + Rs 99 + Rs 404 + Rs 165 = Rs 720.50

Question 15.
Rajeeta purchased a set of cosmetics. She paid Rs 165 for it including VAT. If the rate of VAT is 10%, find the sale price of the set.
Solution:
Total price of set including VAT = Rs 165
Rate of VAT = 10%

Question 16.
Sunita purchases a bicycle for Rs 660. She has paid a VAT of 10%. Find the list price of bicycle.
Solution:
Cost price of bicycle including VAT = Rs. 660
Rate of VAT = 10%

Question 17.
The sales price of a television inclusive of VAT is Rs 13,500. If VAT is charged at the rate of 8% of the list price, find the list price of the television.
Solution:
Sale price of television including VAT = Rs 13,500
Rate of VAT = 8%

Question 18.
Shikha purchased a car with a marked price of Rs 2,10,000 at a discount of 5%. If VAT is charged at the rate of 10%, find the amount Shikha had paid for purchasing the car.
Solution:
Marked price of car = Rs 2,10,000
Rate of discount = 5%

Question 19.
Sharuti bought a set of cosmetic items for Rs 345 including 15% value added tax and a purse for Rs 110 including 10% VAT. What percent is the VAT charged on the whole transactions ?
Solution:
Cost price of set = Rs 345
VAT = 15%

Question 20.
List price of a cooler is Rs 2,563. The rate of VAT is 10%. The customer requests the shopkeeper to allow a discount in the price of the cooler to such an extent that the price remains Rs 2,563 inclusive of VAT. Find the discount in the price of the cooler.
Solution:
List price of cooler = Rs. 2,563
On request the price of cooler is paid Rs 2,563 including VAT
VAT = 10%

Amount of discount = Rs 2,563 – Rs 2,330 = Rs 233

Question 21.
List price of a washing machine is Rs 9,000. If the dealer allows a discount of 5% on the cash payment, how much money will a customer pay to the dealer in cash, if the rate of VAT is 10%.
Solution:
List price of washing machine = Rs 9,000
Rate of discount = 5%
Amount after giving discount

Exercise 9.1

Solve each of the following equations and also verify your solution :
Question 1.
914 = y – 113
Solution:

Question 2.
5×3 + 25 = 1
Solution:

Question 3.
x2 + x3 + x4 = 13
Solution:

Question 4.
x2 + x8 = 18
Solution:

Question 5.
2×3 – 3×8 = 712
Solution:

Question 6.
(x + 2) (x + 3) + (x – 3) (x – 2) – 2x (x + 1) = 0
Solution:
(x + 2) (x + 3) + (x – 3) (x – 2) – 2x (x + 1) = 0
⇒ [x² + (2 + 3) x + 2 x 3] + [x² + (-3 – 2) x + (-3) (-2)] – 2x² – 2x = 0
⇒ x² + 5x + 6 + x² – 5x + 6 – 2x² – 2x = 0
⇒ x² + x² – 2x² + 5x – 5x – 2x + 6 + 6 = 0
⇒ -2x + 12 = 0
Subtracting 12 from both sides,
-2x + 12 – 12 = 0 – 12
⇒ -2x = -12
Dividing by -2,
x = 6
Verification:
L.H.S. = (x + 2) (x + 3) + (x – 3) (x – 2) – 2x (x + 1)
= (6 + 2) (6 + 3) + (6 – 3) (6 – 2) – 2 x 6 (6 + 1)
= 8 x 9 + 3 x 4 – 12 x 7
= 72 + 12 – 84
= 84 – 84
= 0
= R.H.S.

Question 7.
x2 – 45 + x5 +3×10 = 15
Solution:

Question 8.
7x + 35 = 110
Solution:

Question 9.
2x–13 – 6x–25 = 13
Solution:

Question 10.
13 (y – 4) – 3 (y – 9) – 5 (y + 4) = 0
Solution:
13 (y – 4) – 3 (y – 9) – 5 (y + 4) = 0
⇒ 13y – 52 – 3y + 27 – 5y – 20 = 0
⇒ 13y – 3y – 5y – 52 + 27 – 20 = 0
⇒ 13y – 8y – 72 + 27 = 0
⇒ 5y – 45 = 0
Dividing by 5,
y = 9
Verification:
L.H.S. = 13 (y – 4) – 3 (y – 9) – 5 (y + 4)
= 13 (9 – 4) – 3 (9 – 9) – 5 (9 + 4)
= 13 x 5 – 3 x 0 – 5 x 13
= 65 – 0 – 65
= 0
= R.H.S.

Question 11.
23 (x – 5) – 14 (x – 2) = 92
Solution:

Exercise 9.2

Solve each of the following equations and also check your result in each case :
Question 1.
2x+53 = 3x – 10
Solution:
2x+53 = 3x–101
By cross multiplication
⇒ 2x + 5 = 3 (3x – 10)
⇒ 2x + 5 = 9x – 30
⇒ 5 + 30 = 9x – 2x (By transposition)
⇒ 35 = 7x
⇒ x = 5

Question 2.
a–83 = a–32
Solution:

Question 3.
7y+25 = 6y–511
Solution:

Question 4.
x – 2x + 2 – 163 x + 5 = 3 – 72 x.
Solution:

Question 5.
12 x + 7x – 6 = 7x + 14
Solution:

Question 6.
34 x + 4x = 78 + 6x – 6
Solution:

Question 7.
72 x – 52 x = 203 x + 10
Solution:
72 x – 52 x = 203 x + 10

Question 8.
6x+12 + 1 = 7x–33
Solution:

Question 9.
3a–23 + 2a+32 = a + 76
Solution:

Question 10.
x – x–12 = 1 – x–23
Solution:

Question 11.
3×4 – x–12 = x–23
Solution:

Question 12.
5×3 – x–14 = x–35
Solution:
5×3 – x–14 = x–35

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
0.18 (5x – 4) = 0.5x + 0.8
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
Solution:
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
⇒ (9x² + 6x – 24x – 16) – (8x² + 4x – 22x – 11) = x² + 7x – 3x – 21
⇒ 9x² + 6x – 24x – 16 – 8x² – 4x + 22x + 11 = x² + 4x – 21
⇒ 9x² – 8x² – x² + 6x – 24x + 22x – 4x – 4x = -21 + 16 – 11
⇒ 28x – 32x = -32 + 16
⇒ -4x = -16
⇒ x = 4
Verification:
L.H.S. = (3x – 8) (3x + 2) – (4x – 11) (2x + 1)
= (3 x 4 – 8) (3 x 4 + 2) – (4 x 4 – 11) (2 x 4 + 1)
= (12 – 8) (12 + 2) – (16 – 11) (8 + 1)
= 4 x 14 – 5 x 9 = 56 – 45 = 11
R.H.S. = (x – 3) (x + 7) = (4 – 3) (4 + 7) = 1 x 11 = 11
L.H.S. = R.H.S.

Question 25.
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
Solution:
[(2x + 3) + (x + 5)]² + [(2x + 3) – (x + 5)]² = 10x² + 92
⇒ (2x + 3 + x + 5)² + (2x + 3 – x – 5)² = 10x² + 92
⇒ (3x + 8)² + (x – 2)² = 10x² + 92
⇒ 9x² + 2 x 3x x 8 + 64 + x² – 2 x x x 2 + 4 = 10x² + 92
⇒ 9x² + 48x + 64 + x² – 4x + 4 = 10x² + 92
⇒ 9x² + x² – 10x² + 48x – 4x = 92 – 64 – 4
⇒ 44x = 24

Exercise 9.3

Solve the following equations and verify your answer :
Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.

Solution:

Question 22.

Solution:

Question 23.

Solution:

Question 24.
Find a positive value of x for which the given equation is satisfied.

Solution:

Exercise 9.4

Question 1.
Four-fifth of a number is more than three-fourth of the number by 4. Find the number.
Solution:
Let the required number = x
Then four-fifth of the number = 45x
and three- fourth =  34x

Question 2.
The difference between the squares of two consecutive numbers is 31. Find the numbers.
Solution:
Let first number = x
There second number = x + 1
∴ According to the condition :
(x + 1)² – (x)² = 31
⇒ x² + 2x + 1 – x² = 31
⇒ 2x = 31 – 1 = 30 30
⇒ x =  302 = 15
∴  First number = 15
and second number = 15 + 1 = 16
Hence numbers are 15, 16
Check : (16)² – (15)² = 256 – 225 = 31
Which is given
∴  Our answer is correct.

Question 3.
Find a number whose double is 45 greater than its half.
Solution:
Let the required number = x
Double of it = 2x
and half of it =  x2
According to the condition :

Question 4.
Find a number such that when 5 is subtracted from 5 times the number, the result is 4 more than twice the number.
Solution:
Let the required number = x 5
times of it = 5x
twice of it = 2x
According to the condition :
5x – 5 = 2x + 4
⇒ 5x – 2x = 4 + 5
⇒ 3x = 9
⇒ x =93   = 3
Required number = 3
Check :3 x 5-5 = 2×3+4
⇒  15-5 = 6 + 4
⇒ 10= 10
Which is true. Therefore our answer is correct.

Question 5.
A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.
Solution:
Let the number = x
Then fifth part increased by 5 = x5 + 5
Fourth part diminished by 5 = x4  – 5
According to the condition :

Question 6.
A number consists of two digits whose sum is 9. If 27 is subtracted from the number, its digits are reversed. Find the number.
Solution:
Sum of two digits = 9
Let units digit = x
Then tens digit = 9 – x
and number = 10 (9 – x) + x
= 90 – 10x + x = 90 -9x
On reversing the digits,
Units digit = 9 -x tens digit = x
and number = 10 (x) + 9 – x
= 10x + 9- x = 9x + 9
According to the condition :
90 – 9x – 27 = 9x + 9
⇒ 9x + 9x = 90 – 27-9
⇒ 18x = 90- 36 = 54
⇒ x =5418 = 3
Number = 90 – 9x = 90 – 9 x 3 = 90 – 27 = 63
Check : 63 – 27 = 36 (Whose digits are reversed)
Which is true. Therefore our answer is correct.

Question 7.
Divide 184 into two parts such that one- third of one part may exceed one seventh of another part by 8.
Solution:
Sum of two parts = 184
Let first part = x
Then second part = 184 – x
According to the condition :

Question 8.
The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to 23 . What is the original fraction equal to ?
Solution:
Let denominator of the original fraction = x
Then numerator = x – 6
and fraction = x−6x
According to the condition :

Question 9.
A sum of Rs. 800 is in the form of denominations of Rs. 10 and Rs. 20. If the total number of notes be 50, find the number of notes of each type.
Solution:
Total amount = Rs. 800
Total number of notes = 50
Let number of notes of Rs. 10 = x
Then number of notes of Rs. 20 = 50 – x
According to the condition, x x 10 + (50-x) x 20 = 800
⇒  10x + 1000 – 20x = 800
⇒  -10x = 800- 1000 = -200
⇒ x =   −200−10 = 20
∴ Number of 10-rupees notes = 20
and number of 20-rupees notes = 50-20 = 30
Check : 20 x 10 + 30 x 20
= 200 + 600 = 800
Which is true. Therefore our answer is correct.

Question 10.
Seeta Devi has Rs. 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty-five paise coins as she has fifty-paise coins. How many coins of each kind does she have ?
Solution:
Total amount = Rs. 9
Let fifty paise coins = x
Then twenty-five paise coins = 2x
According to the condition :

Question 11.
Sunita is twice as old as Ashima. If six years is subtracted from Ashima’s age and four years added to Sunita’s age, then Sunita will be four times Ashima’s age. How old were they two years ago ?
Solution:
Let age of Ashima = x
Then age of Sunita = 2x
According to the condition :
4 (x – 6) = 2x + 4
⇒  4x-24 = 2x + 4
⇒ 4x-2x = 4 + 24
⇒  2x = 28
⇒ x = 282 = 14
∴  Sunita’s present age = 2x = 2 x 14 = 28 years
and Ashima’s age = 14 years
Two years ago,
Age of Sunita = 28 – 2 = 26 years
and age of Ashima =14-2 = 12 years

Question 12.
The ages of Sonu and Monu are in the ratio 7 : 5. Ten years hence, the ratio of their ages will be 9 : 7. Find their present ages.
Solution:
Ratio in the present ages of Sonu and Monu = 7:5
Let age of Sonu = 7x years
and age of Monu = 5x years
10 years hence,
the age of Sonu = 7x + 10 years
and age of Monu = 5a + 10 years
According to the condition :

Question 13.
Five years ago a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.
Solution:
5 years ago,
Let age of son = x years
Then, age of father = 7a years
Present age of son = x + 5 years
and age of father = 7x + 5 years
5 years hence,
age of son = x + 5 + 5= x+10
and age of father = 7x + 5 + 5 = 7x + 10
According to the condition :
7x + 10 = 3 (x + 10)
⇒  7x + 10 = 3x + 30
⇒  7x -3x= 30 – 10 = 20

Question 14.
I am currently 5 times as old as my son. In 6 years time I will be three times as old as he will be then. What are our ages now ?
Solution:
Let present age of my son = x years
Then my age = 5x years
After 6 years,
my age will be = 5x + 6
and my son’s age = x + 6
According to the condition
5x + 6 = 3 (x + 6)
⇒ 5x+ 6 = 3x+ 18
⇒ 5x – 3x = 18 – 6 ⇒ 2x = 12
⇒ x = 6
∴ Present my age = 5x = 5 x 6 = 30 years
and my son’s age = 6 years

Question 15.
I have Rs. 1000 in ten and five rupees notes. If the number of ten rupees notes that I have is ten more than the number of five rupees notes, how many notes do I have in each denomination ?
Solution:
Total amount = Rs. 1000
Let the number of five rupee notes = x
∴ Ten rupees notes = x + 10
According to the condition,
(x + 10) x 10 + 5 x x x = 1000
⇒ 10a + 100 + 5a = 1000
⇒  15a = 1000- 100 = 900
⇒ x = 90015 = 60
∴  Number of five rupees notes = 60
and number of ten rupees notes = 60 + 10 = 70

Question 16.
At a party, colas, squash anjd fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two fifths drank juice and just three did not drink any thing. How many guests were in all ?
Solution:
Let total number of guests = x
Guests who drank colas = x4
Guests who drank squash = x3
Guests who drank juice = 25 x
Guest who drank none of these = 3
According to the condition :

Question 17.
There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly ?
Solution:
Number of total questions = 180
Let the candidate answers questions correctly = x
∴ Uncorrect or unattended questions =180 -x
total score he got = 450
According to the condition
x x 4-(180-x) x 1 =450
⇒ 4x – 180 + x = 450
⇒ 5x = 450+ 180 = 630
⇒ x =6305 = 126
Number of question which answered correctly = 126

Question 18.
A labourer is engaged for 20 days on the condition that he will receive Rs. 60 for each day, he works and he will be fined Rs. 5 for each day, he is absent, If he receives Rs. 745 in all, for how many days he remained absent ?
Solution:
Total number of days = 20
Let number of days he worked = x
Then number of days he remained absent = 20 – x
According to the condition :
x x 60 – (20 – x) x 5 = 745
⇒  60x- 100 + 5x = 745
⇒  65x = 745 + 100 = 845
⇒  x = 84565 = 13
∴ Number of days he worked =13 days
and number of days he remained absent = 20 – 13 = 7 days.

Question 19.
Ravish has three boxes whose total weight is 60 12 kg. Box B weighs 312 kg more it than A and box C weighs 513 kg more than box B. Find the weight of box A.
Solution:
Total weight of three boxes = 6012 kg.
Let weight of box A = x kg.

Question 20.
The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 12. Find the rational number.
Solution:
Let denominator of the given rational number = x
Then numerator = x – 3
∴ Rational number =x–3x
According to the condition :

Question 21.
In a rational number, twice the numerator is 2 more than the denominator. If 3 is added to each, the numerator and the denominator, the new fraction is 23 . Find the original number.
Solution:

Question 22.
The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/ hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.
Solution:
Distance between two stations = 340 km.
Let the speed of the first train = x km/hr.
Then speed of second train = (x + 5) km/h.
Time = 2 hours
Distance travelled by the first train in 2 hours = 2x km
and distance travelled by the second train = 2 (x + 5) km
According to the condition,

Question 23.
A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/hr, find the speed of the steamer in still water and the distance between the ports.
Solution:
Time taken by a steamer downstream = 9 hours
and upstream = 10 hours Speed of steamer = 1 km/hr.
Let speed of the steamer = x km/h.
According to the condition :
9 (x + 1) = 10 (x – 1)
9x + 9 = 10x – 10 ⇒ 10x – 9x = 9 + 10
⇒ x = 19
∴  Speed of steamer in still water =19 km/h
and distance between two ports = 9 (a + 1) = 9 (19 + 1) = 9 x 20 = 180 km.

Question 24.
Bhagwanti inherited Rs. 12000.00 She invested part of it as 10% and the rest at 12%. Her annual income from these investments is Rs. 1280.00. How much did she invest at each rate ?
Solution:
Total investment = Rs. 12000.00
Rate of interest for first part = 10%
and for second part = 12%
Annual income = Rs. 1280.00
Let the investment for the first part = Rs. x
and second part = Rs. (12000 – x)
According to the condition :

Question 25.
Total investment = Rs. 12000.00 Rate of interest for first part = 10% and for second part = 12% Annual income = Rs. 1280.00 Let the investment for the first part = Rs. a and second part = Rs. (12000 – a) According to the condition :
Solution:
Let breadth of the rectangle = x cm
Then length = (x + 9) cm
∴ Area = length x breadth = x (x + 9) cm²
By increasing each length and breadth by 3 cm
The new length of the rectangle = x + 9 + 3
= (x + 12) cm
and breadth = (x + 3) cm
∴  Area = (x + 12) (x + 3)
According to the condition :
(x + 12) (x + 3) – a (x + 9) = 84
x² + 3x + 12x + 36 – x² – 9x = 84
⇒ 6a = 84 – 36 = 48 ⇒ x  = 486 =8
∴  Length of the rectangle = a + 9 = 8 + 9 = 17 cm
and breadth =x = 8 cm.

Question 26.
The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now ?
Solution:
Sum of ages of Anup and his father =100 years
Let present age of Anup = x years
∴  Age of his father = (100 – x) years
∴  Age of Anuj = 100–x5 years
and also Anuj’s age = (x + 8) years ….I
Anup becomes as old as his father is now
after (100 – 2x) years
∴ After (100 – 2x) years

Question 27.
A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a begger waiting outside the shop. She spent half of what was left on a lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with ?
Solution:
Let the amount, a lady has in the beginning = Rs. x

Exercise 8.1

Question 1.
Write the degree of each of the following polynomials :
(i) 2x³+5x²-7
(ii) 5x² – 3x +7 ’
(iii) 2x + x² –-8
(iv) 12 y⁷ -12y⁵ + 48y⁶ – 10
(v) 3x³ + 1
(vi) 5
(vii) 20x³ + 12x²y²– 10y² + 20
Solution:
(i) 2x³ + 5x²-7: The degree of this polynomial is 3.
(ii) 5x² – 3x + 2 : The degree of this polynomial is 2.
(iii) 2x + x² – 8 : The degree of this polynomial is 2.
(iv) 12 y⁷ – 12y⁶ + 48y⁵ – 10 : The degree of this polynomial is 7.
(v) 3x³ + 1 : The degree of this polynomial is 3.
(vi) 5 : The degree of this polynomial is 0 as it is only constant term
(vii) 20x³ + 12x²y² – 10y² + 20: The degree of this polynomial is 2 + 2 = 4.

Question 2.
Which of the following expressions are not polynomials :
(i) x² + 2x²                   
(ii) √a x + x²-x³
(iii) 3y³ – √5y + 9      
(iv) ax^1/2 + ax + 9x² + 4
(v) 3x² + 2x^-1 + 4x + 5
Solution:
(i) x² + 2x^-2 = x² + 2x 1×2 =x² + 1×2
: It is not xx polynomial as it has negative integral power.
(ii) √ax + x² – x³: It is polynomial.
(iii) 3y³  √5y + 9 : It is a polynomial.
(iv) ax^1/2+ ax + 9x² + 4: It is not a polynomial as  the degree of 1×2 is an integer.
(v) 3x² + 2x^-1 + 4x + 5 : It is not a polynomial as the degree of x^–2, x^-1 are negative.

Question 3.
Write each of the following polynomials in the standard form. Also write their degree.
(i) x² + 3 + 6x + 5x⁴
(ii) a¹ + 4 + 5a⁶
(iii) (x³ – 1) (x³ – 4)
(iv) (y³ – 2) (y³ + 11)
(v) (a3−38) (a3−1617)
(vi) (a+34) (a+34)
Solution:
Polynomial in standard form is the polynomial in ascending order or descending order.

Exercise 8.2

Question 1.
6x³y²z² by 3x²yz
Solution:

Question 2.
15m²n³ by 5m²n²
Solution:

Question 3.
24a³b³ by -8ab
Solution:

Question 4.
-21abc² by 7abc
Solution:

Question 5.
72xyz² by – 9xz
Solution:

Question 6.
-72a⁴b⁵c⁸ by – 9a²b²c³
Solution:

Simplify :

Question 7.

Solution:

Question 8.

Solution:

Exercise 8.3

Question 1.
x+2x²+3x⁴-x⁵ by 2x
Solution:

Question 2.
y⁴-3y³+ 12 y² by 3y
Solution:

Question 3.
-4a³ + 4a² + a by 2a
Solution:

Question 4.
-x⁶ + 2x⁴ + 4.x³ + 2x² by √2x²
Solution:

Question 5.
5z³ – 6z² + 7z by 2z
Solution:

Question 6.
√3 a⁴ + 2 √3 a³ + 3a² – 6a by 3a
Solution:

Exercise 8.4

Question 1.
5x³ – 15x² + 25x by 5x
Solution:

Question 2.
4z³ + 6z²-zby −12 z
Solution:

Question 3.
9x²y – 6xy + 12xy² by −32 xy
Solution:

Question 4.
3x²y² + 2x²y + 15xy by 3xy
Solution:

Question 5.
x² + 7x + 12 by x + 4
Solution:

Question 6.
4y² 4 + 3y + −12 by 2y + 1
Solution:

Question 7.
3x³ + 4x² + 5x + 18 by x + 2
Solution:

Question 8.
14x² – 53x + 45 by 7a – 9
Solution:

Question 9.
-21 + 71x – 31x² – 24ax³ by 3 – 8ax
Solution:

Question 10.
3y⁴ – 3y³ – 4y² – 4y by y² – 2y
Solution:

Question 11.
2y⁵ + 10y⁴ + 6y³ + y² + 5y + 3 by 2y³ + 1
Solution:

Question 12.
x⁴ – 2x³ + 2x² + x + 4 by x² + x + 1
Solution:

Question 13.
m³ – 14m² + 37m – 26 by m² – 12m + 13
Solution:

Question 14.
x⁴ + x² + 1 by x² + x + 1
Solution:

Question 15.
x⁵ + x⁴ + x³+x² + x+ 1 by x³ + 1
Solution:

Divide each of the following and find the quotient and remainder :

Question 16.
14x³ – 5x² + 9x -1 by 2x – 1
Solution:

Question 17.
6x³ – x² – 10x – 3 by 2x – 3
Solution:

Question 18.
6x³+ 11x² – 39x – 65 by 3x² + 13x + 13
Solution:

Question 19.
30a⁴ + 11a³-82a²– 12a + 48 by 3a² + 2a- 4
Solution:

Question 20.
9x⁴ – 4x² + 4 by 3x² – 4x + 2
Solution:

Question 21.
Verify division algorithm i.e., Dividend = Divisor * Quotient + Remainder, in each of the following. Also, write the quotient and remainder :

Solution:

Question 22.
Divide 15y⁴ + 16y³ + 103 y – 9y² – 6 by 3y – 2
Write down the co-efficients of the terms in the quotient.
Solution:

Question 23.
Using division of polynomials state whether.
(i) x + 6 is a factor of x² – x – 42
(ii) 4x – 1 is a factor of 4x² – 13x – 12
(iii) 2y – 5 is a factor of 4y⁴ – 10y³ – 10y² + 30y -15
(iv) 3y + 5 is a factor of 6y⁵ + 15y + 16y + 4y+ 10y – 35
(v) z² + 3 is a factor of z⁵– 9z
(vi) 2x² – x + 3 is a factor of 60x⁵-x⁴ + 4x³ – 5x² -x- 15
Solution:

Question 24.
Find the value of ‘a’, if x + 2 is a factor of 4x⁴ + 2x³ – 3x² + 8x + 5a.
Solution:

Question 25.
What must be added to x⁴ + 2x³ — 2x² + x – 1 so that the resulting polynomial is exactly divisible by x² + 2x – 3.
Solution:

Exercise 8.5

Question 1.
Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder.
(i) 3x² + 4x + 5, x – 2
(ii) 10x² – 7x + 8, 5x – 3
(iii) 5y³– 6y² + 6y-1,5y-1
(iv)x⁴-x³ + 5x,x-1
(v) y⁴ +y²,y²-2
Solution:
(i) 3x² + 4x + 5, x – 2
= 3x (x – 2) + 10x + 5
= 3x (x – 2) + 10 (x – 2) + 25
∴ Quotient = 3x + 10
Remainder = 25

(iii) 5y³ – 6y² + 6y – 1, 5y – 1
= y² (5y – 1) – 5y² + 6y- 1
= y² (5y – 1) -y (5y – 1) + 5y – 1
= y² (5y- 1) -y (5y- 1) + 1 (5y- 1)
∴ Quotient = y² – y + 1 and Remainder = 0
(iv) x⁴ – x³ + 5x, x – 1
= x³(x – 1) + 5x
= x³ (x – 1) + 5 (x – 1) + 5
∴ Quotient = x³ + 5, Remainder = 5
(v) y⁴+y²,y²– 2
= y²(y² – 2) + 3y²
= y² (y² – 2) + 3 (y² – 2) + 6
∴ Quotient =y² + 3 and Remainder = 6

Question 2.
Find, whether or not the first polynomial is a factor of the second :
(i) x + 1, 2x² + 5x + 4
(ii) y- 2, 3y³ + 5y² + 5y + 2
(iii) 4x² – 5, 4.x⁴ + 7x² + 15
(iv) 4-z, 3z² – 13z + 4
(v) 2a-3,10a² – 9a – 5
(vi) 4y+1 ,8y²-2y + 1
Solution:
(i) x + 1, 2x² + 5x + 4
2x² + 5x + 4 = 2x (x + 1) + 3x + 4
= 2x (x + 1) + 3 (x + 1) + 1
∵ Remainder = 1
∴ x + 1 is not a factor of 2x² + 5x + 4
(ii) y – 2, 3y³ + 5y² + 5y + 2
3y³ + 5y² + 5y + 2 = 3y²(y – 2)+11y² + 5y + 2
= 3y²(y – 2)+11y (y – 2) + 27y + 2
= 3y² (y – 2) + 11y (y – 2) + 27 (y – 2) + 56
∵ Remainder = 56
∴ y – 2 is not a factor of 3y³ + 5y² + 5y + 2
(iii) 4x² – 5, 4x⁴ + 7x² + 15
4x⁴ + 7x² + 15 = x² (4x² – 5) + 12x² + 15
= x² (4x² – 5) + 3 (4x² – 5) + 30
∵ Remainder = 30
∴ 4x² – 5 is not a factor of 4x⁴ + 7x² + 15
(iv) 4 – z, 3z² – 13z + 4
3z² – 13z + 4 = -3z (-z + 4) – z + 4
= -3z (-z + 4) + 1 (-z + 4)
∵ Remainder = 0
∴ 4 – z or – z + 4 is a factor of 3z² – 13z + 4
(v) 2a – 3, 10a² – 9a – 5
10a² – 9a – 5 = 5a (2a – 3) + 6a – 5
= 5a (2a – 3) + 3 (2a – 3) + 4
∵ Remainder = 4
∴ 2a – 3 is not a factor of 10a² – 9a – 5
(vi) 4y + 1, 8y² – 2y + 1
8y² – 2y + 1 = 2y (4y + 1) – 4y + 1
= 2y (4y + 1) – 1 (4y + 1) + 2
∵ Remainder = 2
∴ 4y + 1 is not a factor of 8y² – 2y + 1

Exercise 8.6

Divide :

Question 1.
x² – 5x + 6 by x – 3
Solution:

Question 2.
ax² – ay² by ax + ay
Solution:

Question 3.
x⁴ – y⁴ by x² – y²
Solution:

Question 4.
acx² + (bc + ad)x + bd by (ax + b)
Solution:

Question 5.
(a² + 2ab + b²)- (a² + 2ac + c²) by 2a + b + c
Solution:

Question 6.
14 x² – 12 x- 12 by 12 x – 4
Solution:

Exercise 7.1

Find the greatest common factors (GCF / HCF) of the following polynomials : (1 – 14)

Question 1.
2x² and 12x²
Solution:
2x² and 12x²
HCF of 2 and 12 =2
HCF of x²,x²=x²
∴ HCF = 2x²

Question 2.
(6xy³ and 18x²y³
Solution:
6x³y and 18xy
HCF of 6, 18 = 6
HCF of x³ and x² = x²
HCF of y and y³ -y
∴ HCF = 6x²y

Question 3.
7x, 21x² and 14xy²
Solution:
7x, 21x² and 14xy²
HCF of 7, 21 and 14 = 7
HCF of x, x², x = x
∴ HCF = 7x

Question 4.
42x²yz and 63x³y²z³
Solution:
42x²yz and 63x³y²z³
HCF of 42 and 63 = 21
HCF of x², x³ = x²
HCF of y,y²=y
HCF of z,z³ = z
∴ HCF = 21 x²yz

Question 5.
12ax²,6a²x³ and 2a³ x⁵
Solution:
12ax², 6a²x³ and 2a³x⁵
HCF of 12, 6,2 = 2
HCF of a, a², a³ = a
HCF of x², x³, x⁵ = x²
∴ HCF = 2ax²

Question 6.
9x², 15x²y³, 6xy² and 21x²y²
Solution:
9x², 15xV, 6xy² and 21x²y²
HCF of 9, 15, 6,21 = 3
HCF of x², x², x, x² = x
HCF of 1, y³, y², y² =2
∴ HCF = 3x

Question 7.
4a²b³ -12a³b, 18a⁴b³
Solution:
4a²b³, -12a³b, 18a⁴b³
HCF of 4, 12, 18 = 2
HCF of a², a³, a⁴ = a²
HCF of b³,b, b³ = b
∴ HCF = 2a²b

Question 8.
6x²y², 9xy³, 3x³y²
Solution:
6x²y², 9xy³, 3x³y²
HCF of 6, 9, 3 = 3
HCF of x², x, x³ = x
HCF of y²,y³,y²=y²
∴ HCF = 3xy²

Question 9.
a²b³, a³b²
Solution:
a²b³, a³b²
HCF of a², a³ = a²
HCF of b³, b² = b²
∴ HCF = a²b²

Question 10.
36a²b²c⁴, 54a⁵c²,90a⁴b²c²
Solution:
36a²b²c⁴, 54a⁵c²,90a⁴b²c²
HCF of 36, 54, 90 = 18
HCF of a², a⁵, a⁴ = a²
HCF of b², 1,b²= 1
HCF of c⁴,c²,c² = c²
∴ HCF = 18a² x 1 x c² = 18a²c²

Question 11.
x³, – yx²
Solution:
x³, – yx²
HCF of x³, x² = x²
HCF of 1, y= 1
∴ HCF = x²

Question 12.
15a³, -45a², -150a
Solution:
15a³,-45a²,-150a
HCF of 15,45, 150 = 15
HCF of a³, a², a = a
∴ HCF = 15a

Question 13.
2x³y², 10x²y³, 14xy
Solution:
2x³y², 10x²y³, 14xy
HCF of 2, 10, 14 = 2
HCF of x³, x², x = x
HCF of y²,y³,y=y
∴ HCF = 2xy

Question 14.
14x³y⁵, 10x⁵y³, 2x²y²
Solution:
14x³y⁵, 10x⁵y³, 2x²y²
HCF of 14, 10, 2, = 2
HCF of x³, x⁵, x² = x²
HCF of y⁵,y³,y²=y²
∴ HCF = 2xy

Find the greatest common factor of the terms in each of the following expressions:

Question 15.
5a⁴ + 10a³ – 15a²
Solution:
5a⁴ + 10a³– 15a²
HCF of 5, 10, 15 = 5
HCF of a⁴, a³, a² = a²
∴ HCF = 5a²

Question 16.
2xyz + 3x²y + 4y²
Solution:
2xyz + 3x²y + 4y²
HCF of 2, 3,4 = 1
HCF of x, x², 1 = 1
HCF of y,y,y² =y
HCF of z, 1, 1 = 1
∴ HCF = y

Question 17.
3a²b² + 4b²c² + 12a²b²c²
Solution:
3a²b² + 4b²c² + 12a²b²c²
HCF of 3, 4, 12 = 1
HCF of a², 1, a² = 1
HCF of b², b², b² = b²
HCF of 1, c², c² = 1
∴ HCF = b²

Exercise 7.2

Factorize the following :

Question 1.
3x-9
Solution:
3x – 9 = 3 (x – 3)        (HCF of 3, 9 = 3)

Question 2.
5x – 15x²
Solution:
5x- 15x² = 5x (1 – 3x)
{HCF of 5, 15 = 5 and of x, x² = x}

Question 3.
20a¹²b² – 15a⁸b⁴
Solution:
20a¹²b² – 15a⁸b⁴
{HCF of 20, 15 = 5, a¹², a⁸ = a⁸, b², b⁴ = b²}
= 5a⁸ b²(4a⁴ – 3b²)

Question 4.
72xy – 96x⁷y⁶
Solution:
72xy – 96x⁷y⁶
HCF of 72, 96 = 24 of x⁶x⁷ = x⁶, y⁷,y⁶ = y⁶
∴ 72x⁷y⁶ – 96x⁷y⁶ = 24x⁶y⁶ (3y – 4x)

Question 5.
20X³ – 40x² + 80x
Solution:
20x³ – 40x² + 80x
HCF of 20, 40,80 = 20
HCF of x³, x², x = x
∴ 20x³ – 40x² + 80x = 20x (x² – 2x + 4)

Question 6.
2x³y² – 4x²y³ + 8xy⁴
Solution:
2x³y² – 4x²y³ + 8xy⁴
HCF of 2, 4, 8 = 2
HCF of x³, x², x = 1
and HCF of y², y³, y⁴ = y²
∴ 2x³y² – 4x²y³ + 8xy⁴
= 2xy² (x² – 2xy + 4y²)

Question 7.
10m³n² + 15m⁴n – 20m²n³
Solution:
10m³n² + 15m⁴n – 20m²n³
HCF of 10, 15, 20 = 5
HCF of m³, m⁴, m² = m²
HCF of n², n, n³ = n
10m³n² + 15m⁴n – 20m²n³
5m²n(2mn + 3m²– 4n²)

Question 8.
2a⁴b⁴ – 3a³b⁵ + 4a²b⁵
Solution:
2a⁴b⁴ – 3a³b⁵ + 4a²b⁵
HCF of 2, 3, 4= 1
HCF of a⁴, a³, a² = a²
HCF of b⁴, b⁵ b⁵ = b⁴
∴ 2a⁴b⁴ – 3a³b⁵ + 4a²b⁵ = a²b⁴
(2a² – 3ab + 4b)

Question 9.
28a² + 14a²b² – 21a⁴
Solution:
28a² + 14a²b² – 21a⁴
HCF of 28, 14,21 =7
HCF of a², a², a⁴ = a²
HCF of 1, b², 1 = 1
∴ 28a² + 14a²b²-21a⁴ = 7a²
(4 + 2b² – 3a²)

Question 10.
a⁴b – 3a²b² – 6ab³
Solution:
a⁴b – 3a²b² – 6ab³
HCF of 1,3,6 = 1
HCF of a⁴, a², a = a
HCF of b, b², b³ = b
∴ a⁴b – 3a²b² – 6ab³ = ab (a³ – 3ab – 6b²)

Question 11.
2l²mn – 3lm²n + 4lmn²
Solution:
2l²mn – 3lm²n + 4lmn²
HCF 2, 3,4 = 1,
HCF of l²,l,l = l
HCF of m, m², m = m
HCF of n, n, n² = n
∴ 2l² mn – 3lm²n + 4lmn²
= lmn (21 -3m + 4n)

Question 12.
x⁴y² – x²y⁴ – x⁴y⁴
Solution:
x⁴y² – x²y⁴ – x⁴y⁴
HCF of x⁴, x², x⁴ = x²
HCF of y², y⁴, y⁴ =y²
∴ x⁴y² – x²y⁴ – x⁴y⁴ = x²y² (x² -y² -x²y²)

Question 13.
9 x²y + 3 axy
Solution:
9 x²y + 3 axy
HCF of 9, 3 = 3
HCF of x², x = x
HCF of y,y = y
HCF of 1,a = 1
∴ 9x²y + 3axy = 3xy (3x + a)

Question 14.
16m – 4m²
Solution:
16m – 4m²
HCF of 16, 4 = 4
HCF of m, m² = m
∴ 16m – 4m² = 4m (4 – m)

Question 15.
-4a² + 4ab – 4ca
Solution:
-4a² + 4ab – 4ca
HCF of 4, 4, 4 = 4
HCF of a², a, a = a
∴ -4a² + 4ab – 4ca = -4a (a – b + c)

Question 16.
^{x2yz + xy2z + xyz2}
Solution:
x²yz + xy²z + xyz²
HCF of x², x, x = x
HCF of y,y²,y=y
HCF of z, z,z² = z
∴ x²yz + xy²z + xyz² = xyz (x + y + z)

Question 17.
ax²y + bxy² + cxyz
Solution:
ax²y + bxy² + cxyz
HCF of x², x, x = x,
HCF of y,y²,y = y
ax²y + bxy² + cxyz = xy (ax + by + cz)

Exercise7.3

Factorize each of the following algebraic expressions.

Question 1.
6x (2x – y) + 7y (2x – y)
Solution:
6x (2x – y) + 7y (2x – y)
= (2x – y) (6x + 7y)
[∵ (2x – y) is common]

Question 2.
2r (y – x) + s (x – y)
Solution:
2r (y – x) + s (x – y)
-2r (x – y) +s (x – y)
= (x – y) (-2r + s)                   [(x – y) is common]
= (x-y) (s-2r)

Question 3.
7a (2x – 3) + 2b (2x – 3)
Solution:
7a (2x – 3) + 3b (2x – 3)
= (2x – 3) (7a + 3b)               [(2x – 3) is common]

Question 4.
9a (6a – 5b) – 12a² (6a – 5b)
Solution:
9a (6a – 5b) – 12a² (6a – 5b)
HCF of 9 and 12 = 3
∴ 3a (6a – 5b) (3 – 4a)
{(6a – 5b) is common}

Question 5.
5 (x – 2y)² + 3 (x – 2y)
Solution:
5 (x – 2y)² + 3 (x – 2y)
= 5 (x – 2y) (x – 2y) + 3 (x – 2y)
= (x – 2y) {5 (x – 2y) + 3}
{(x – 2y) is common}
= (x – 2y) (5x – 10y + 3)

Question 6.
16 (2l – 3m)² – 12 (3m – 2l)
Solution:
16 (2l – 3m)² – 12 (3m-2l)
= 16 (2l – 3m) (2l – 3m) + 12 (2l – 3m)
HCF of 16, 12 = 4 4 (2l-3m) {4 (2l- 3m) + 3}
{(2l – 3m) is common}
= 4 (2l -3m) (8l- 12m+ 3)

Question 7.
3a (x – 2y) – b (x – 2y)
Solution:
3a (x – 2y) – b (x – 2y)
= (x – 2y) (3a – b)
{(x – 2y) is common}

Question 8.
a² (x + y) + b² (x + y) + c² (x + y)
Solution:
a² (x + y) + b² (x + y) + c² (x + 3’)
= (x + y) (a² + b² + c²)
{(x + y) is common}

Question 9.
(x-y)² + (x -y)
Solution:
(x – y)² + (x- y) = (x – y) (x – y) + (x – y)
= (x – y) (x – y + 1)                          {(a – y) is common}

Question 10.
6 (a + 2b) – 4 (a + 2b)²
Solution:
6 (a + 2b) – 4 (a + 2b)²
= 6 (a + 2b) – 4 (a + 2b) (a + 2b)
HCF of 6, 4 = 2
= 2 {a + 2b) {3 – 2 {a + 2b)
{2 (a + b) is common}
= 2 (a + 2b) (3-2 a- 4b)

Question 11.
a (x -y) + 2b (y – x) + c (x -y)²
Solution:
a (x -y) + 2b (y – x) + c (x -y)²
= a (x – y) – 2b (x – y) + c (x – y) {x – y)
= (x – y) {x – 2b + c (x – y)}
{(a – y) is common}
= (a – y) (a – 2b + cx – cy)

Question 12.
– 4 (a – 2y)² + 8 (a – 2y)
Solution:
– 4 (x – 2y)² + 8 (x – 2y)
= – 4 (x – 2y) (x – 2y) + 8 (x – 2y)
{- 4 (x – 2y) is common}
= – 4 (x – 2y) (x – 2y – 2)
= 4 (x – 2y) (2 – x + 2y)

Question 13.
x³ (a – 2b) + a² (a – 2b)
Solution:
x³ (a – 2b) + x² (a – 2b)
HCF of x³, x² = x^2
∴ x² (a – 2b) (x + 1)
{x² (x – 2b) is common}
= x² (x – 2b) (x + 1)

Question 14.
(2x – 3y) (a + b) + (3x – 2y) (a + b)
Solution:
(2x – 3y) (a + b) + (3x – 2y) (a + b)
= (a + b) {2x – 3y + 3x – 2y}
{(x + b) is common}
= (a + b) (5x – 5y)
= 5 (a + b) (x – y)

Question 15.
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
Solution:
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
= 4 (x + y) (3a – b) – 6 (x + y) (3a – 2b)
HCF of 4, 6 = 2
= 2 (x + y) {2 (3a – b) – 3 (3a – 2b)}
= 2 (x + 3) {6a – 2b – 9a + 6b}
= 2 (x +y) {-3a + 4b}
= 2 (x + y) (4b – 3a)

Exercise 7.4

Factorize each of the following expressions :
Question 1.
qr-pr + qs – ps
Solution:
qr- pr + qs-ps
Arranging in suitable groups = r(q-p) +s (q-p)    {(q – p) is common}
= (q-p) (r + s)

Question 2.
p²q -pr²-pq + r²
Solution:
p²q -pr²-pq + r2
= p²q -pq-pr² + r² (Arranging in group)
= pq(p- 1)-r²(p-1) {(p – 1) is common}
= (p – 1) (pq – r²)

Question 3.
1 + x + xy + x²y
Solution:
1 + x + xy + x 2y
= 1 (1 + x) +xy (1 +x)
= (1 + x) (1 + xy) {(1 + x) is common}

Question 4.
ax + ay – bx – by
Solution:
ax + ay – bx – by
= a (x + y) – b (x + y)   {(x + y) is coinmon}
= (x+y) (a- b)

Question 5.
xa² + xb² -ya² – yb²
Solution:
xa² + xb² – ya² – yb²
= x (a² + b²) -y (a² + b²)   {(a² + b²) is common}
= {a² + b²) (x -y)

Question 6.
x² + xy + xz + yz
Solution:
x² + xy + xz + yz
= x (x + y) + z(x + y) {(x + y) is common}
= (x + y) (x + z)

Question 7.
2ax + bx + 2ay + by
Solution:
2ax + bx + 2ay + by
= x {2a + b) + y (2a + b)      {(2a + b) is common}
= (2a + b) (x + y)

Question 8.
ab- by- ay +y²
Solution:
ab – by – ay + y²
= b(a-y)-y(a-y)    {(a -y) is common}
= (a-y) (b – y)

Question 9.
axy + bcxy -az- bcz
Solution:
axy + bcxy – az – bcz
= xy (a + bc) – z (a + bc)       {(a + bc) is common}
= (a + bc) (xy – z)

Question 10.
lm² – mn² – lm + n²
Solution:
lm² – mn² – lm + n²
= m (lm – n²)- 1 (lm – n²)  {(lm – n²) is common}
= (lm – n²) (m – 1)

Question 11.
x³ – y² + x – x²y²
Solution:
x³ -y² + x – x²y²
⇒ x³ + x – x²y² – y²
= x(x²+ 1)-y²(x²+ 1)        {(x² + 1) is common}
= (x² + 1) (x -y²)

Question 12.
6xy + 6 – 9y – 4x
Solution:
6xy + 6 – 9y – 4x
= 6 xy – 4x – 9y + 6
= 2x (3y – 2) – 3 (3y – 2)    {(3y – 2) is common}
= (3y-2) (2x – 3)

Question 13.
x² – 2ax – 2ab + bx
Solution:
x² – 2ax – 2ab + bx
⇒ x² – 2ax + bx – 2ab
= x (x – 2a) + b (x – 2a)   {(x – 2a) is common}
= (x – 2a) (x + b)

Question 14.
x³ – 2x²y + 3xy² – 6y³
Solution:
x³ – 2x²y + 3xy² – 6y³
= x² (x – 2y) + 3y² (x – 2y)     {(x – 2y) is common}
= (x – 2y) (x² + 3y²)

Question 15.
abx² + (ay – b) x-y
Solution:
abx² + (ay – b) x-y
= abx² + ayx – bx -y 
= ax (bx + y) – 1 (bx + y)               {(bx +y) is common}
= (bx + y) (ax – 1)

Question 16.
(ax + by)² + (bx – ay)²
Solution:
(ax + by)² + (bx – ay)²
= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy
= a²x² + b²y² + b²x² + a²y²
= a²x² + b²x² + a²y² + by²
= x² (a² + b²) + y² (a² + b²)         {(a² + b²) is common}
= (a² + b²) (x² + y²)

Question 17.
16 (a – b)³ -24 (a- b)²
Solution:
16 (a – b)³ -24 (a- b)²
HCF of 16, 24 = 8
and HCF of (a – b)³, (a – b)² = (a – b)²
∴16 (a – b)³ – 24 (a – b)²
= 8 (a-b)² {2 (a-b)- 3}
{8 (a – b)² is common}
= 8 (a – b)² (2a – 2b – 3)

Question 18.
ab (x² + 1) + x (a² + b²)
Solution:
ab (x² + 1) + x (a² + b²)
= abx² + ab + a²x + b²x
= abx² + b²x + a²x + ab
= bx (ax + b) + a (ax + b)  {(ax + b) is common}
= (ax + b) (bx + a)

Question 19.
a²x² + (ax² + 1) x + a
Solution:
a²x² + (ax² + 1) x + a
= a²x² + ax³ + x + a
= ax³ + a²x² + x + a
= ax² (x + a) + 1 (x + a) {(x + a) is common}
= (x + a) (ax² + 1)

Question 20.
a(a- 2b -c) + 2bc
Solution:
a(a- 2b -c) + 2bc
= a²– 2ab -ac +2bc
= a (a – 2b) – c (a – 2b) {(a – 2b) is common}
= (a – 2b) (a – c)

Question 21.
a (a + b – c)- bc
Solution:
a (a + b – c) – bc
= a² + ab – ac – bc
= a (a + b) – c (a + b)   {(a + b) is common}
= (a + b) (a – c)

Question 22.
x² – 11xy – x +11y
Solution:
x² – 11xy-x + 11y
= x² -x – 11 xy + 11 y
= x (x – 1) – 11y (x – 1)   {(x – 1) is common}
= (x- 1) (x- 11y)

Question 23.
ab – a – b + 1
Solution:
ab – a-b + 1
= a (b – 1) – 1 (b – 1)    {(b – 1) is common}
= (b – 1) (a – 1)

Question 24.
x² + y – xy – x
Solution:
x² + y – xy – x
= x² – x- xy + y
= x (x – 1) – y (x – 1)   {(x – 1) is common}
= (x- 1) (x-y)

Exercise 7.5

Factorize each of the following expressions :
Question 1.
16x²-25y²
Solution:
16x² – 25y² = (4x)² – (5y)²    {∵ a² – b² = (a + b) (a – b)}
= (4x + 5y) (4x – 5y)

Question 2.
27x² – 12y²
Solution:
27x² – 12y² = 3 (9x² – 4y²)  {∵ a² -b² = (a + b) (a – b)}
= 3 [(3x)² – (2y)²]
= 3 (3x + 2y) (3x – 2y)

Question 3.
144a² – 289b²
Solution:
144a² – 289b² = (12a)² – (17b)²    { ∵ a² – b² = (a + b) (a – b}
= (12a+ 17b) (12a- 17b)

Question 4.
12m² – 27
Solution:
12m² – 27 = 3 (4m² – 9)
= 3 {(2m)²-(3)²}   {∵ a² – b² = (a + b) (a – b)}
= 3 (2m + 3) (2m – 3)

Question 5.
125x² – 45y²
Solution:
125x² – 45y² = 5 (25x² – 9y²)
= 5 {(5x-)² – (3y)²}    {∵ a² – b² = (a + b) (a – b}
= 5 (5x + 3y) (5x – 3y)

Question 6.
144a² – 169b²
Solution:
144a² – 169b² = (12a)² – (13b)²    {∵ a² -b² = (a + b) (a – b)}
= (12a + 13b) (12a-13b)

Question 7.
(2a – b)² – 16c²
Solution:
(2a – b)² – 16c² = (2a – b)² – (4c)²   {∵ a² – b² = (a + b) (a – b)}
= (2a – b + 4c) (2a – b – 4c)

Question 8.
(x + 2y)² – 4 (2x -y)²
Solution:
(x + 2y)² – 4 (2x – y)²
= (x + 2y)² – {2 (2x –y)}²
= (x + 2y)² – (4x – 2y)²        {∵ a²– b² = (a + b) (a – b)}
= (a + 2y + 4x – 2y) (x + 2y – 4x + 2y)
= 5x (-3x + 4y)

Question 9.
3a⁵ – 48a³
Solution:
3a⁵ – 48a³ = 3a³ (a²– 16)
= 3a³ {(a)² – (4)²}        {∵ a² – b² = (a + b) (a – b)}
= 3a³ (a + 4) (a – 4)

Question 10.
a⁴ – 16b⁴
Solution:
a⁴ – 16b⁴ = (a²)² – (4b²)²
= (a² + 4b²) (a² – 4b²)
= (a² + 4b²) {(a)² – (2b)² }   { ∵ a² – b² = (a + b) (a – b)}
= (a² + 4b²) (a + 2b) (a – 2b)

Question 11.
x⁸ – 1
Solution:
x⁸ – 1 = (x⁴)² – (1)²
= (x⁴ + 1) (x⁴ – 1)
= (x⁴+ 1) I (x²)² – (1)²}             {∵ a² – b² = (a + b) (a – b)}
= (x⁴ + 1) (x² + 1) (x² – 1)
= (x⁴ + 1) (x² + 1) {(x)² – (1)²}
= (x⁴+ 1)(x² + 1)(x+ 1)(x- 1)
= (x-1)(x+ 1) (x² + 1) (x⁴ + 1)

Question 12.
64 – (a + 1)²
Solution:
64 – (a + 1)² = (8)² – (a + 1)²    {∵ a² – b² = (a + b) (a – b)}
= (8 + a + 1) (8 – a – 1)
= (9 + a) (7 – a)

Question 13.
36l² – (m + n)²
Solution:
36l² – (m + n)² = (6l)² – (m + n)²        {∵  a² – b² = (a + b) (a – b)}
= (6l + m + n) (6l – m – n)

Question 14.
25x⁴y⁴ – 1
Solution:
25x⁴y⁴ – 1 = (5x⁴y⁴)² – (1)²         { ∵  a² – b² = (a + b) (a – b)}
= (5x⁴y⁴  + 1) (5x²y²  – 1)

Question 15.

Solution:

Question 16.
x³ – 144x
Solution:
x³ – 144x = x (x² – 144)
= x {(x)² – (12)²}       {∵ a² – b² = (a + b) (a – b)}
=  x (x + 12) (x – 12)

Question 17.
(x – 4y)² – 625
Solution:
(x – 4y)² – 625
= (x – 4y)² – (25)²     {∵ a² – b² = (a + b) (a – b)}
= (x – 4y + 25) (x -4y – 25)

Question 18.
9 (a – b)² – 100 (x -y)²
Solution:
9(a-b)²– 100(x-y)²
= {3(a-b)}²-{10(x-y)}²      {∵ a² – b² = (a + b) (a – b)}
= (3a – 3b)² – (10x – 10y)²
= (3a – 3b + 10x – 10y) (3a – 3b – 10x + 10y)

Question 19.
(3 + 2a)² – 25a²
Solution:
(3 + 2a)² – 25a²
= (3 + 2a)² – (5a)²      (∵ a² – b² = (a + b) (a – b)}
= (3 + 2a + 5a) (3 + 2a – 5a)
= (3 + 7a) (3 – 3a)
= (3 + 7a) 3 (1 – a)
= 3(1-a) (3 +7a)

Question 20.
(x + y)² – (a – b)²
Solution:

Question 21.

Solution:

Question 22.
75a³b² – 108ab⁴
Solution:
75a³b² – 108ab⁴
= 3ab² (25a² – 36b²)
= 3ab² {(5a)² – (6b)²}         {∵ a² – b² = (a + b) (a – b)}
= 3ab² (5a + 6b) (5a – 6b)

Question 23.
x⁵– 16x³
Solution:
x⁵ – 16x³ = x³ (x² – 16)
= x³ {(x)² – (4)²} {∵ a² – b² = (a + b) (a – b)}
= x³ (x + 4) (x – 4)

Question 24.

Solution:

Question 25.
256x⁵ – 81x
Solution:
256x⁵– 81x = x(256x⁴– 81)
= x {(16x²)² – (9)²}      {∵ a² – b² = {a + b) (a – b)}
= x (16x² + 9) (16x² – 9)
= x (16x² + 9) {(4x)² – (3)²}
= x (16x² + 9) (4x + 3) (4x-3)

Question 26.
a⁴ – (2b + c)⁴
Solution:
a⁴ – (2b + c)⁴
= (a²)² – [(2b + c)²]²    {∵ a² – b² = (a + b) (a – b)}
= {a² + (2b + c)²} {a² – (2b + c)²}
= {a² + (2b + c)²} {(a)² – (2b + c)²}
= {a² + (2b + c)²} (a + 2b + c) (a -2b- c)

Question 27.
(3x + 4y)⁴ – x⁴
Solution:
(3x + 4y)⁴ – x⁴ – [(3x + 4y)²]² – (x²)²
= [(3x + 4y)² + x²] [(3x + 4y)² – x²]       {∵  a² – b² = (a + b) (a – b)
= [(3x + 4y)² + x²] [(3x + 4y + x) (3x + 4y – x)]
=   [(3x + 4y)² + x²] (4x + 4y) (2x + 4y)
= [(3x + 4y)² + x²] 4 (x + y) 2 (x + 2y)
= 8 (x + y) (x + 2y) [(3x + 4y)² + x²]

Question 28.
p²q² – p⁴q⁴
Solution:
p²q²– p⁴q⁴ =p²q² (1 -p²q²)
=p²q² [(1)² – (pq)²]   {∵ a² – b² = (a + b) (a – b)
= p²q² (1 +pq) (1 -pq)

Question 29.
3x³y – 243xy³
Solution:
3x³y – 243xy³
= 3xy (x² – 81y²)
= 3xy [(x)² – (9y)²]
= 3xy (x + 9y) (x – 9y)

Question 30.
a⁴b⁴ – 16c⁴
Solution:
a⁴b⁴ – 16c⁴ = (a²b²)² – (4c²)²
= (a²b² + 4c²) (a²b² – 4c²)
= (a²b² + 4c²) [(ab)² – (2c)²]      {∵ a² – b² = (a + b) (a – b)
= (a²b² + 4c²) (ab + 2c) (ab – 2c)

Question 31.
x⁴-625
Solution:
x⁴ – 625 = (x²)² – (25)²   {∵ a² – b² – (a + b) (a – b)
= (x² + 25) (x² – 25)
= (x² + 25) [(x)² – (5)²]
= (x² + 25) (x + 5) (x – 5)

Question 32.
x⁴-1
Solution:
x⁴ – 1 = (x²)² – (1)² = (x² + 1) (x² – 1)
= (x² + 1) [(x)² – (1)²]
= (x² + 1) (x + 1) (x – 1)

Question 33.
49 (a – b)² -25 (a + b)²
Solution:
49 (a – by -25 (a + b)²
= [7 (a – b)]² – [5 (a + b)]²
= (7a – 7b)² – (5a + 5b)²  {∵ a² – b² = (a + b) (a – b)
= (7a -7b + 5a + 5b) (7a – 7b -5a- 5b)
=(12a – 2b)(2a – 12b)
= 2 (6a – b) 2 (a – 6b)
= 4 (6 a- b) (a – 6b)

Question 34.
x – y – x² + y² 
Solution:
x-y-x² + y² = (x-y)-(x²-y²) {∵ a² – b² = (a + b) (a – b)
= {x-y)-(x + y)(x-y)
= (x-y)(1 – x – y)

Question 35.
16 (2x – 1)² – 25y²
Solution:
16 (2x – 1)² – 25y²
= [4 (2x – 1)]² – (5y)²
= (8x – 4)² – (5y)²
= (8x – 4 + 5y) (8x -4-5y)
= (8x + 5y – 4) (8x – 5y – 4)

Question 36.
4 (xy + 1)² – 9 (x – 1)²
Solution:
4 (xy + 1)² – 9 (x – 1)^2
= [2 (xy + 1)]² – [3 (x – 1)]²
= (2xy + 2)² – (3x – 3)² {∵ a² – b² = (a + b) (a – b)
= (2xy + 2 + 3x – 3) (2xy + 2 – 3x + 3)
= (2xy + 3x – 1) (2xy – 3x + 5)

Question 37.
(2x + 1)² – 9x⁴
Solution:
(2x + 1)² – 9x⁴ = (2x + 1)² – (3x²)²    {∵ a² – b² = (a + b) (a – b)
= (2x + 1 + 3x²) (2x + 1 – 3x²)
= (3x² + 2x + 1) (-3x + 2x + 1)

Question 38.
x⁴ – (2y- 3z)²
Solution:
x⁴ – (2y – 3z)² = (x²)² – (2y – 3z)²
= (x² + 2y- 3z) (x² – 2y + 3z)

Question 39.
a²-b² +a-b
Solution:
a² – b² + a – b
= (a + b) {a – b) + 1 (a – b)
= (a – b) (a + b + 1)

Question 40.
16a⁴ – b⁴
Solution:
16a⁴ – b⁴
= (4a²)² – (b²)²            { ∵  a² – b² = (a + b) (a – b)
= (4a² + b²) (4a² – b²)
= (4a² + b²) {(2a)² – (b)²}
= (4a² + b²) (2a + b) (2a – b)

Question 41.
a⁴ – 16 (b – c)⁴
Solution:
a⁴ – 16 (b- c)⁴ = (a²)² – [4 (b – c)²]² { ∵  a² – b² = (a + b) (a – b)
= [a² + 4 (b – c)²] [a² – 4 (b – c)²]
= [a² + 4 (b – c)²] [(a)² – [2 (b – c)]²]
= [a² + 4 (b – c)²] [(a)² – (2b – 2c)²]
= [a² + 4 (b – c)²] (a + 2b – 2c) (a – 2b + 2c)

Question 42.
2a⁵ – 32a
Solution:
2a⁵ – 32a = 2a (a⁴ – 16)
= 2a [(a²)² – (4)²]  {∵  a² – b² = (a + b) (a – b)
= 2a (a² + 4) (a² – 4)]
= 2a (a² + 4) [(a)² – (2)²]
= 2a (a² + 4) (a + 2) (a – 2)

Question 43.
a⁴b⁴ – 81c⁴
Solution:
a⁴b⁴ – 81c⁴ = (a²b²)² – (9c²)²
= (a²b² + 9c²) (a²b² – 9c²) {∵ a² – b² = (a + b) (a – b)
= (a²b² + 9c²) {(ab)² – (3c)²}
= (a²b² + 9c²) (ab + 3c) (ab – 3c)

Question 44.
xy⁹-yx⁹
Solution:
xy⁹ – yx⁹ = xy (y⁸ – x⁸)
= xy [(y⁴)² – (x⁴)²] {∵  a² – b² = (a + b) (a – b)}
= xy(y⁴ + x⁴)(y⁴-x⁴)
= xy (y⁴ + x⁴) {(y²)² – (x²)²}
= xy (y⁴ + x⁴) (y² + x²) (y² – x²)
= xy (y⁴ + x⁴) (y² + x²) (y + x) (y – x)

Question 45.
x³ -x
Solution:
x³-x = x(x²– 1)
= x [(x)² – (1)²] = x (x + 1) (x – 1)

Question 46.
18a²x² – 32
Solution:
18a²x² – 32
= 2 [9a²x² – 16]
= 2 [(3ax)² – (4)²]   {∵ a² – b² = (a + b) (a – b)
= 2 (3ax + 4) (3ax – 4)

Exercise 7.6

Factorize each of the following algebraic expressions :
Question 1.
4x² + 12xy + 9y²
Solution:
4x² + 12xy + 9y² = (2x)² + 2 x 2x x 3y + (3y)² {∵ a² + 2ab + b² = (a +b)²}
= (2x + 3y)²    

Question 2.
9a² – 24ab + 16b²
Solution:
9a² – 24ab + 16b²
= (3a)² – 2 x 3a x 4b + (4b)²     {∵ a² – 2ab + b² = (a – b)²}
= (3a – 4b)²

Question 3.
36a² – 6pqr + 9r²
Solution:
p²q² – 6pqr + 9r²
= (pq)² – 2 x pq x3r + (3r)² {∵ a² – 2ab + b² = (a -b)²}
= (pq-3r)²

Question 4.
36a² + 36a + 9
Solution:
36a² + 36a + 9
= (6a)² + 2 x 6a x 3 + (3)²   {∵ a² + 2ab + b² = (a + b)²
= (6a + 3)²

Question 5.
a² + 2ab + b² – 16
Solution:
a² + 2ab + b² – 16
= (a + b)² – (4)²     {∵ a² + 2ab + b² = (a + b)² and a² – b² = (a + b) (a – b)}
= (a + b + 4) (a + b – 4)

Question 6.
9z² – x² + 4xy – 4y²
Solution:
9z² – x² + 4xy – 4y²    {∵ a² – b² = (a + b) (a – b) and a² – 2ab + b² (a – b)²}
= 9z² – (x² – 4xy + 4y²)
= (3z)² – [(x)² – 2 x x x 2y + (2y)²]
= (3z)²-(x-2y)²
= (3z + x – 2y) (3z – x + 2y)

Question 7.
9a⁴ – 24a²b² + 16b⁴ – 256
Solution:

Question 8.
16 – a⁶ + 4a³b³ – 4b⁶
Solution:

Question 9.
a² – 2ab + b² – c²
Solution:

Question 10.
x² + 2x + 1 – 9y²
Solution:

Question 11.
a² + 4ab + 3b²
Solution:
a² + 4ab + 3b²
= a² + 4ab+ 4b² – b²
= (a)² + 2 x a x 2b + (2b)² – b² (∵ 3b² = 4b² – b²)
= (a + 2b)² – (b)²   {∵ a² – b² = (a +b) (a – b)}
= (a + 2b + b) (a + 2b- b)

Question 12.
96 – 4x-x²
Solution:
96 – 4x – x² = 96 – (4x + x²)
= 96 – [(x)² + 2 x x x 2 + (2)²] + (2)²   (on completing the square)
= 96 + 4 – (x + 2)² = 100 – (x + 2)²
= (10)² – (x + 2)²
= (10 + x + 2) (10 – x- 2)
= (x + 12) (-x + 8)

Question 13.
a⁴ + 3a² + 4
Solution:
a⁴ + 3a² + 4
= (a²)² + (2)² + 2 x a² x 2 – a²   (on completing the square)
= (a² + 2)² – (a)²
= (a² + 2 + a) (a² + 2 – a)
= (a¹ + a + 2) (a² – a + 2)

Question 14.
4a⁴ + 1
Solution:
4x⁴ + 1 = (2a²)² + (1)² + 2 x 2x² x 1 – 2 x 2x² x 1  (completing the square)
= (2x² + 1)² – 4a²
= (2x² + 1)² – (2a)²   {a² – b² = (a + b) (a – b)}
= (2x² + 1 + 2a) (2a² + 1 – 2a)
= (2a² + 2a + 1) (2a² – 2a + 1)

Question 15.
4x⁴+y⁴
Solution:
4a⁴ + y⁴ = (2x²)² + (y²)² + 2 x 2x²y² – 2 x 2x²y²
= (2x² + y²)² – 4x²y²
= (2x² + y²)² – (2xy)²
= (2x² + y² + 2xy) (2x² + y² – 2xy)
= (2x² + 2xy + y²) (2x² – 2xy + y²)

Question 16.
(x+ 2)² – 6 (a + 2) + 9
Solution:
(x + 2)² – 6 (x + 2) + 9
=  (x + 2)² – 2 x (x + 2) x 3 + (3)²
= (x + 2 – 3)²
= (x-1)² = (x-1)(x-1)

Question 17.
25 – p² – q² – 2pq
Solution:

Question 18.
a² + 9y² – 6xy – 25a²
Solution:
a² + 9y² – 6xy – 25a²

Question 19.
49 – a² + 8ab – 16b²
Solution:

Question 20.
a² – 8ab + 16b² – 25c²
Solution:

Question 21.
x² -y²+ 6y- 9
Solution:

Question 22.
25x² – 10x + 1 – 36y²
Solution:

Question 23.
a²-b² + 2bc – c²
Solution:

Question 24.
a² + 2ab + b² -c²
Solution:

Question 25.
49 -x² – y² + 2xy
Solution:

Question 26.
a² + 4b² – 4ab – 4c²
Solution:

Question 27.
x² -y² – 4xz + 4z²
Solution:

Exercise 7.7

Factorize each of the following algebraic expressions :
Question 1.
x² + 12x – 45
Solution:

Question 2.
40 + 3x – x²
Solution:

Question 3.
a² + 3a-88
Solution:

Question 4.
a² – 14a – 51
Solution:

Question 5.
x² + 14x + 45
Solution:

Question 6.
x² – 22x + 120
Solution:

Question 7.
x²– 11x – 42
Solution:

Question 8.
a² + 2a – 3
Solution:

Question 9.
a² + 14a + 48
Solution:

Question 10.
x² – 4x – 21
Solution:

Question 11.
y² – 5y-36
Solution:

Question 12.
(a²-5a)²-36
Solution:

Question 13.
(a + 7) (a – 10) + 16
Solution:

Exercise 7.8

Resolve each of the following quadratic trinomials into factors :
Question 1.
2x² + 5x + 3
Solution:

Question 2.
2x²– 3x – 2
Solution:

Question 3.
3x² + 10x + 3
Solution:

Question 4.
7x – 6 – 2x²
Solution:

Question 5.
7x² – 19x – 6
Solution:

Question 6.
28-31x -5x²
Solution:

Question 7.
3 + 23y – 8y²
Solution:

Question 8.
11x² – 54x + 63
Solution:

Question 9.
7x-6x² + 20
Solution:

Question 10.
3x² + 22x + 35
Solution:

Question 11.
12x² – 17xy + 6y²
Solution:

Question 12.
6x² – 5xy – 6y²
Solution:

Question 13.
6x² + 13xy + 2y²
Solution:

Question 14.
14x² + 11xy – 15y²
Solution:

Question 15.
6a² + 17ab – 3b²
Solution:

Question 16.
36a² + 12abc – 15b²c²
Solution:

Question 17.
15x² – 16xyz – 15y²z²
Solution:

Question 18.
(x – 2y)² -5 (x- 2y) + 6
Solution:

Question 19.
(2a – b)² + 2 (2a – b) – 8
Solution:

Exercise 7.9

Factorize each of the following quadratic polynomials by using the method of completing the square.
Question 1.
p² + 6p + 8
Solution:
p² + 6p + 8
= p² + 2 x p x 3 + 3² – 3² + 8   (completing the square)
= (p² + 6p + 3²) – 1
= (p + 3)² – 1²
= (P + 3)² – (1)²       { ∵ a² + b²  = (a+b) (a-b)}
= (p +3+1) (p + 3 -1)
= (p+4) (p+ 2)

Question 2.
q² – 10q + 21
Solution:
q² – 10q + 21
= (q)² – 2 x q x 5 + (5)² – (5)² + 21   (completing the square)
= (q)² – 2 x q x 5 + (5)² -25+21
= (q)²-2 x q x 5 + (5)² – 25 +21
= (q)²-2 x q x 5 + (5)² – 4
= (q – 5)² – (2)     {∵ a² – b² = (a + b) (a – b)}
= (q- 5 + 2) (q-5-2)
=(q- 3) (q-7)

Question 3.
4y² + 12y + 5
Solution:
4y² +12y + 5
= (2y)² + 2 x 2y x 3 + (3)² – (3)² + 5    (completing the square)
= (2y + 3)² – 9 + 5
= (2y + 3)² – 4
= (2y + 3)²-(2)²   {∵ a² – b² = (a + b) (a – b)}
= (2y + 3 + 2) (2y + 3 – 2)
= (2y + 5) (2y+ 1)

Question 4.
p² + 6p- 16
Solution:
p² + 6p – 16
= (p)² + 2 x  p x 3 + (3)² – (3)² – 16    (completing the square)
= (p)² + 2 x p x 3 + (3)² – 9 – 16
= (p + 3)² – 25
= (p + 3)² – (5)²     {∵ a² -b² = {a + b) (a – b)}
= (p + 3 + 5)(p + 3-5)
= (p + 8) (p – 2)

Question 5.
x² + 12x + 20
Solution:
x² + 12x + 20
= (x)² + 2 x x x 6 + (6)² – (6)² + 20   (completing the square)
= (x)² + 2 x x x6 + (6)² -36 + 20
= (x + 6)² -16
= (x + 6)² – (4)²   {∵ a² – b² = (a + b) (a – b)}
= (x + 6 + 4) (x + 6 – 4)
= (x + 10) (x + 2)

Question 6.
a² – 14a – 51
Solution:
a² – 14a-51
= (a)² – 2 x x 7 + (7)² – (7)² – 51       (completing the square)
= (a)² – 2 x a x 7 + (7)² – 49 – 51
= (a – 7)² – 100
= (a – 7)² – (10)²    {∵  a² – b² = (a + b) (a – b)}
= (a – 7 + 10) (a – 7 – 10)
= (a + 3) (a – 17)

Question 7.
a² + 2a – 3
Solution:
a² + 2a – 3
= (a)² + 2 x a x 1 + (1)² – (1)2 – 3   (completing the square)
= (a)² + 2 x a x 1 + (1)² – 1 – 3
= (a + 1)² – 4
= (a + 1)² – (2)² {∵ a² – b² = (a + b) (a – b)}
= (a + 1 + 2) (a + 1 – 2)
= (a + 3) (a – 1)

Question 8.
4x² – 12x + 5
Solution:
4x² – 12x + 5
= (2x)² – 2 x 2x x 3 + (3)² – (3)² + 5  (completing the square)
= (2x)² – 2 x 2x x 3 + (3)² -9 + 5
= (2x – 3)² – 4
= (2x – 3)² – (2)²      {∵ a² – b² = (a + b) (a – b)}
= (2x – 3 + 2) (2x – 3 – 2)
= (2x – 1) (2x – 5)

Question 9.
y² – 7y + 12
Solution:

Question 10.
z²-4z-12
Solution:
z² – 4z – 12
= (z)² – 2 x z x 2 + (2)² – (2)² – 12  (completing the square)
= (z)² – 2 x z x 2 + (2)² – 4 – 12
= (z-2)²-16
= (z-2)²-(4)²   {∵ a² – b² = (a + b) (a – b)}
= (z – 2 + 4) (z – 2 – 4)
= (z + 2)(z-6)

Exercise 6.1

Question 1.
Identify the terms, their co-efficients for each of the following expressions.
(i) 7x²yz – 5xy
(ii) x² + x + 1
(iii)3x²y² – 5x²y² + z² + z²
(iv) 9 – ab + be-ca
(v) a2+b2-ab
(vi)2x – 0.3xy + 0.5y
Solution:
(i) Co-efficient of 7x²yz = 7
co-efficient of -5xy = -5
(ii) Co-efficient of x¹ = 1
co-efficient of x = 1
co-efficient of 1 = 1
(iii) Co-efficient of 3x²_y² = 3
co-efficient of -5x²y²z² = -5
co-efficient of z² – 1
(iv) Co-efficient of 9 = 9
co-efficient of -ab = -1
co-efficient of be = 1
co-efficient of -ca = -1
(v) Co-efficient of a2=12
Co-efficient of b2=12
co-efficient of -ab = -1
(vi) co-efficient of 0.2x = 0.2
co-efficient of-0.3xy = -0.3
co-efficient of 0.5y = 0.5

Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category ?
(i) x+y
(ii) 1000
(iii) x + x² + x³ + x⁴
(iv) 7 + a + 5b
(v) 2b – 3 b²    
(vi) 2y – 3y² +4y³
(vii) 5x – 4y + 3x
(viii) 4a – 15a²
(ix) xy+yz + zt + tx
(x)   pqr
(xi) p²q + pq²       
(xii)  2p + 2 q
Solution:
Monomials are (ii), (x)
Binomials are (i), (v), (viii), (xi), (xii)
Trinomials are (iv), (vi) and (vii)
None of these are (iii) and (ix)

Exercise 6.2

Question 1.
Add the following algebraic expressions


Solution:

Question 2.
Subtract:
(i) -5xy from 12xy
(ii) 2a² from -7a²

Solution:

Question 3.
Take away :


Solution:

Question 4.
Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and – 4X + 9y- 11z.
Solution:
Sum of x – 3y + 2z and – 4x + 9y – 11z
= x – 3y + 2z + (- 4x + 9y – 11z)
= x – 3y + 2z – 4x + 9y – 11z
= x – 4x – 3y + 9y + 2z – 11z
= – 3x + 6y – 9z
Now (-3x + 6y – 9z) – (3x – 4y – 7z)
= -3x + 6y – 9z – 3x + 4y + 7z
= -3x – 3x + 6y + 4y -9z +7z
= -6x + 10y – 2z

Question 5.
Subtract the sum of 3l- 4m – 7n² and 2l + 3m – 4n² from the sum of 9l + 2m – 3n² and -3l + m + 4n².
Solution:
Sum of 9l + 2m – 3n² and -3l + m + 4n²
= 9l + 2m – 3 n² + (-3l) + m + 4n²
= 9l + 2m – 3n² – 3l + m + 4n²
= 9l- 3l+ 2m + m – 3 n² + 4n²
= 6l + 3m + n²
and sum of 3l – 4m – 7n² and 2l +3m- 4n²
= 3l- 4m – 7n² + 2l+ 3m- 4n²
= 3l + 2l – 4m + 3m- 7n² – 4n²
= 5l -m- 11n²
Now (6l + 3m + n²) – (5l – m – 11n²)
= 6l + 3m + n² – 5l + m + 11n²
= 6l – 5l + 3m + m + n² + 11n²
= l + 4m+ 12n²

Question 6.
Subtract the sum of 2x – x² + 5 and -4x – 3 + 7x² from 5.
Solution:
5 – (2x-x² + 5-4x-3 + 7x²)
= 5 – (2x – 4x- x² + 7x² + 5-3)
= 5 – (-2x + 6x² + 2)
= 5 + 2x – 6x² – 2
= – 6x²+2x+3
= 3 + 2x – 6x²

Question 7.

Solution:

Exercise 6.3

Find each of the following products (1-8)
Question 1.
5x² x 4x³
Solution:
5x² x 4x³ = 5 x 4 x x² x x³
= 20x^{2 + 3} = 20x^s

Question 2.
3a² x 4b⁴
Solution:
-3a² x 4b⁴ = -3 x 4 x a²b⁴
= -12a²b⁴

Question 3.
(-5xy) x (-3x²yz)
Solution:
(-5xy) x (-3x²yz)
= (-5) x (-3)xy x x²yz
= 15x^{1 + 2}xy^{1+ 1}z= 15x³y²z

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Find each of the following products : (9-17)

Question 9.
(7ab) x (-5ab²c) x (6abc²)
Solution:
(7ab) x (-5ab²c) x (6abc²)
= 7 x (-5) x 6 x a x a x a x b x b² x b x c x c²
=-210 x a¹⁺¹⁺¹ x b¹⁺²⁺¹x c¹⁺²
=-210 x a³b⁴c³

Question 10.
(-5a) x (-10a²) x (-2a³)
Solution:
(-5a) x (-10a²) x (-2a³)
= (-5) (-10) (-2) x a x a² x a³
= -100a^{1 + 2 + 3} = -100a⁶

Question 11.
(-4x²) x (-6xy²) x (-3yz²)
Solution:
(-4x²) x (-6xy²) x (-3yz²)
= (-4) x (-6) x (-3) x² x x x y² x y xz²
= -72x²⁺¹ x y²⁺¹ x _z²
= 72x³y³z³

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.
(2.3xy) x (0.1x) x (0.16)
Solution:
(2.3xy) x (0.1x) x (0.16)
= 2.3 x 0.1 x 0.16 x x x x x y
= 0.0368x^{1 +1} x y = 0.0368x²y

Express each of the following products as a monomials and verify the result in each case for x = 1 : (18 -26)

Question 18.
(3x) x (4x) x (-5x)
Solution:

Question 19.

Solution:

Question 20.
(5x⁴) x (x²)³ x (2x)²
Solution:
(5x⁴) x (x²)³ x (2x)²
= 5x⁴ x x^{2 x 3} x 2x x 2x
= 5x⁴ * x⁶ x 4x² = 5 x 4 x x^{4 + 6 + 2}
= 20x¹²
Verification:
L.H.S. = (5x⁴) x (x²)³ x (2x)²
= 5 x (1)⁴ x [(1)²]³ x (2 x 1)²
= 5 x 1 x (1)^{2 x 3}_x (2)²
= 5 x 1⁶ x 2² = 5 x 1 x 4 = 20
R.H.S. = 20x¹² = 20 (1)¹² = 20 x 1 = 20
∴ L.H.S. = R.H.S.

Question 21.
(x²)³ x (2x) x (-4x) x 5
Solution:
(x²)³ x (2x) x (-4x) x (5)
= x^{2 x 3} X 2x X (-4x) X 5
= x⁶ x 2x x (-4x) x 5 = 2 x (-4) x 5x^{6+1 +1}
= -40x⁸
Verification
L.H.S. = (x²)³ x (2x) x (-4x) x (5)
= (1²)³ x (2 x 1) x (-4 x 1) x 5
= 1^2 x 3 x 2 x (- 4) x 5 = 1⁶ x 2 x (-4) x 5
= 1 x 2 x (-4) x 5 = -40
R.H.S. = -40x⁸ = -40 x (1)⁸
= -40 x 1 = -40
∴ L.H.S. = R.H.S.

Question 22.
Write down the product of -8x²y⁶ and – 20xy Verify the product for x = 2.5, y = 1.
Solution:
Product of -8x²y⁶ and -20xy
= (-8x²y⁶) x (-20xy)
= -8 x (-20) x² x _x x y⁶ x y = 160x^{2 + 1} x y^{6 + 1}
= 160x³y³
Verification.
L.H.S. = (-8x²y⁶) x (-20xy)
= -8 x (2.5)² x (1) x (-20 x 2.5 x 1)
= -8 x 6.25 x 1 x -20 x 2.5
= (-50) x (-50) = 2500
R.H.S. = 160 x = 160 (2.5)³ x (1)⁷
= 160 x 15.625 x 1 =2500
∴ L.H.S. = R.H.S.

Question 23.
Evaluate : (3.2x⁶y³) x (2.1x²y²) when x = 1 and y = 0.5.
Solution:
3.2x⁶y³ x 2.1x²y²
= 3.2 x 2.1 x x⁶⁺² x y³⁺²
= 6.72x⁸y⁵ = 6.72 x (1)⁸ x (0.5)⁵
= 6.72 x 1 x 0.03125
= 0.21

Question 24.
Find the value of (5x⁶) x (-1.5x²y³) x (-12xy²) when x = 1, y = 0.5.
Solution:

Question 25.
Evaluate : (2.3a⁵b²) x (1.2a²b²) when a = 1, b = 0.5.
Solution:

Question 26.
Evaluate : (-8x²y⁶) x (-20xy) for x = 2.5 and y = 1.
Solution:

Express each of the following products as a monomials and verify the result for x = 1,y = 2: (27-31)

Question 27.
(-xy³) x (yx³) x (xy)
Solution:
(-xy³) x (yx³) x (xy)
= -x x x³ x x x y³ x y x y = -x^{1 + 3 + 1} x y^{3 + 1 + 1} = -x⁵y⁵
Verification:
L.H.S. = (-xy³) x (yx³) x (xy)
= (-1 x 2³) x [2 x (1)³] x (1 X 2)
= (-1 x 8) x (2 x 1) x (1 x 2)
= -8 x 2 x 2 = -32
R.H.S. =-x⁵y⁵  = -(1)⁵ (2)⁵
= -1 x 32 =-32
∴ L.H.S. = R.H.S.

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.

Solution:

Exercise 6.4

Find the following products 

Question 1.
2a³ (3a + 5b)
Solution:
2a³ (3a + 5b) = 2a³ x 3a + 2a³ x 5b
= 6a^{3 +1} + 10a³b
= 6a⁴ + 10a³b

Question 2.
-11a (3a + 2b)
Solution:
-11a (3a + 2b) = -11a x 3a – 11a x 2b
= -33a²– 22ab

Question 3.
-5a (7a – 2b)
Solution:
-5a (7a – 2b) = -5a x 7a- 5a x (-2b)
= -35a² + 10ab

Question 4.
-11y² (3y + 7)
Solution:
-11y² (3y + 7) = -11y² x 3y – 11y² x 7
= -33y²⁺¹-77y²
= 33y³-77y²

Question 5.
6×5 (x³+y³)
Solution:

Question 6.
xy (x³-y³)
Solution:
xy (x³ – y³) =xy x x³ – xy x y³
= x^1 + 3 x y – x x y¹⁺³
= x⁴y – xy⁴

Question 7.
0.1y (0.1x⁵ + 0.1y)
Solution:
0.1y (0.1x⁵ + 0.1y) = 0.1y x 0.1x⁵ + 0.1y x 0.1y
= 0.01x⁵y + 0.01y²

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.
5x (10x²y – 100xy²)
Solution:
5x (10x²y – 100xy²)
= 1.5x x 10x²y – 1.5x x 100xy²
= 15x^{1 + 2}y- 150x¹⁺¹ x y²
15 x³y- 150x² y²

Question 12.
4.1xy (1.1x-y)
Solution:
4.1xy (1.1x-y) = 4.1xy x 1.1x – 4.1xy x y
= 4.51x²y-4.1xy²

Question 13.

Solution:

Question 14.

Solution:

Question 15.
43 a (a² + 6² – 3c²)
Solution:

Question 16.
Find the product 24x² (1 – 2x) and evaluate its value for x = 3.
Solution:
24x² (1 – 2x) = 24x² x 1 + 24x² x (-2x)
= 24x² + (-48x²⁺¹)
= 24x² – 48x³
If x = 3, then
= 24 (3)² – 48 (3)³
= 24 x 9-48 x 27 = 216- 1296
= -1080

Question 17.
Find the product of -3y (xy +y²) and find its value for x = 4, and y = 5.
Solution:
-3y (xy + y²) = -3y x xy – 3y x y²
= -3xy² -3y^{2 +1}  = -3xy² – 3y³
If x = 4, y = 5, then
= -3 x 4 (5)² – 3 (5)³ = -12 x 25 – 3 x 125
= -300 – 375 = – 675

Question 18.
Multiply – 32 x²y³ by (2x-y) and verify the answer for x = 1 and y = 2.
Solution:

Question 19.
Multiply the monomial by the binomial and find the value of each for x = -1, y = 25 and z =05 :
(i) 15y² (2 – 3x)
(ii) -3x (y² + z²)
(iii) z² (x – y)
(iv) xz (x² + y²)
Solution:

Question 20.
Simplify :
(i) 2x² (at¹ – x) – 3x (x⁴ + 2x) -2 (x⁴ – 3x²)
(ii) x³y (x² – 2x) + 2xy (x³ – x⁴)
(iii) 3a² + 2 (a + 2) – 3a (2a + 1)
(iv) x (x + 4) + 3x (2x² – 1) + 4x² + 4
(v) a (b-c) – b (c – a) – c (a – b)
(vi) a (b – c) + b (c – a) + c (a – b)
(vii) 4ab (a – b) – 6a² (b – b²) -3b² (2a² – a) + 2ab (b-a)
(viii) x² (x² + 1) – x³ (x + 1) – x (x³ – x)
(ix) 2a² + 3a (1 – 2a³) + a (a + 1)
(x) a² (2a – 1) + 3a + a³ – 8
(xi) 32-x² (x² – 1) + 14-x² (x² + x) – 34x (x³ – 1)
(xii) a²b (a – b²) + ab² (4ab – 2a²) – a³b (1 – 2b)
(xiii )a²b (a³ – a + 1) – ab (a⁴ – 2a² + 2a) – b (a³– a² -1)
Solution:
(i) 2x² (x³ -x) – 3x (x⁴ + 2x) -2 (x⁴ – 3x²)
= 2x² x x³-2x²x x-3x x x⁴-3x x 2x-2x⁴ + 6x²
= 2x^{2 + 3}– 2x^{2 +1} – 3x^{,1+ 4}-6x^,1+1 -2x⁴ + 6x²
= 2x⁵ – 2x³ – 3x⁵ — 6x² – 2x⁴ + 6x²
= 2x⁵ – 3x⁵ – 2a⁴ – 2x³ + 6x² – 6x²
= -x⁵ – 2x⁴ – 2x³ + 0
= -x⁵-2x⁴-2x³
(ii) x³y (x² – 2x) + 2xy (x³ – x⁴)
= x³y x x² – x³y x 2x + 2ay x ac³ – 2xy x x⁴
= x^{3 + 2}y-2x^{3 + 1} y + 2x^{1 + 3}y – 2yx⁴⁺¹
= x⁵y – 2x⁴y + 2x⁴y – 2yx⁵
= -x⁵y
(iii) 3a² + 2 (a + 2) – 3a (2a + 1)
= 3a² + 2a + 4 – 6a² – 3a
= 3a² – 6a² + 2a – 3a + 4
= -3a² – a + 4
(iv) x (x + 4) + 3x (2x² – 1) + 4x² + 4
= x² + 4x + 3x x 2x² – 3x x 1 + 4x² + 4
= x² + 4x + 6x^{2 +1} – 3x + 4x² + 4
= x² + 4x + 6x³ – 3x + 4x² + 4
= 6a³ + 4x² + x² + 4x – 3x + 4
= 6x³ + 5x² + x + 4
(v) a (b – c)-b (c – a) – c (a – b)
= ab – ac – be + ab – ac + bc
= 2ab – 2ac
(vi) a (b – c) + b (c – a) + c (a – b)
= ab – ac + bc – ab + ac – bc
= ab – ab + bc – be + ac – ac
= 0
(vii) 4ab (a – b) – 6a² (b – b²) -3b² (2a² – a) + 2ab (b – a)
= 4a²b – 4ab² – 6a²b + 6a²b² – 6a²b² + 3ab² + 2ab² – 2a²b
= 4a²b- 6a²b – 2 a²b – 4ab² + 3 ab² + 2ab² + 6a²b² – 6a²b²
= 4a²b – 8a²b – 4ab² + 5 ab² + 0
= – 4a²b + ab²
(viii) x² (x² + 1) – x³ (x + 1) – x (x³ – x)
= x^{2 + 2} + x² – x^{3 + 1} – x³ – x^{1 + 3} + x^{1 + 1}
= x⁴ + x²-x⁴-x³-x⁴ + x²
= x⁴-x⁴-x⁴-x³ + x² + x²
= -x⁴ – x³ + 2x²
(ix) 2a² + 3a (1 – 2a³) + a (a + 1)
= 2a² + 3 a – 3 a x 2a³ + a² + a
= 2a² + 3a – 6a^{1 + 3} + a² + a
= 2a² + 3a – 6a⁴ + a² + a
= -6a⁴ + 3a² + 4a
(x) a² (2a – 1) + 3a + a³ – 8
= 2 a² x a – a² x 1+3a + a³-8
= 2a³ – a² + 3a + a³ – 8
= 2a³ + a³ – a² + 3a – 8
= 3a³ – a² + 3a – 8

(xii) a²b (a – b²) + ab² (4ab – 2a²) – a³b (1 – 2b)
= a²b x a – a²b x b¹ + ab² x 4ab – ab¹ x2a² -a³b x 1 + a³b x 2b
= a²⁺¹ b-a²b^{2 +1}+ 4a^{1 +1} b^{2 +1} -2a²⁺¹ b²-a³b + 2a³b^1 +1
= a’b – a²b³ + 4a²b³ – 2a³b² – a³b + 2a³b²
= a³b – a³b – a²b³ + 4a²b³ – 2a³b² + 2a³b²
= 0 + 3a²b³ + 0 = 3 a²b³
(xiii) a²b (a³ – a + 1) – ab (a⁴ – 2a² + 2a) – b (a³ -a²– 1)
= a²b x _a³ – a²b x _a + a²b – ab x a² + ab x 2a² – ab x 2a- ba³ + ba² + b
= a^2+ 3b – a²⁺¹ b + a²b -a^{1 + 4}b + 2a^{1 + 2}b- 2a¹⁺¹ b- a³b + a²b + b
= a⁵b – a³b + a²6 – a⁵b + 2a³b – 2a²b – a³b + a²b + b
= a⁵b – a³b + 2a³b – a³6 – a³b + a²b – 2a²b + a²b + b
= a³b – a⁵b + 2a³b – 2a³b + 2a²b-2a²b + b
= 0 + 0 + 0 + b = b

Exercise 6.5

Question 1.
(5x + 3) by (7x + 2)
Solution:
(5x + 3) x (7x + 2)
= 5x (7x + 2) + 3 (7x + 2)
= 35x² + 10x + 21x + 6
= 35x² + 31x + 6

Question 2.
(2x + 8) by (x – 3)
Solution:
(2x + 8) x (x – 3)
= 2x (x – 3) + 8 (x – 3)
= 2x² – 6x + 8x – 24
= 2x² + 2x – 24

Question 3.
(7x +y) by (x + 5y)
Solution:
(7x + y) x (x + 5y)
= 7x (x + 5y) + y (x + 5y)
= 7x² + 35xy + xy + 5y²
=7x² + 36xy + 5y²

Question 4.
(a – 1) by (0.1a² + 3)
Solution:
(a – 1) x (0.1a² + 3)
= a (0.1a² + 3) – 1 (0.1a²+ 3)
= 0.1a³ + 3a-0.1a²-3
= 0.1a³ – 0.1a² + 3a-3

Question 5.
(3x² +y²) by (2x² + 3y²)
Solution:
(3x²+y²) x (2x² + 3y²)
= 3x² (2x² + 3y²) + y²(2x² + 3y²)
= 6x^{2 +2} + 9x²y² + 2x²y² + 3y^{2 + 2}
= 6x⁴ + 11 x²y² + 3y⁴

Question 6.

Solution:

Question 7.
(x⁶-y⁶) by (x²+y²)
Solution:
(x⁶ – y⁶) x (x² + y²)
= x⁶ (x² + y²) – y⁶ (x² + y²)
= x⁶ x x² + x⁶y² – x²y⁶ -y⁶ x y²
= x^{6 + 2} + x⁶y² – x²y⁶ – y^6 +2
= x  + x⁶y² – x²y⁶ – y⁸

Question 8.
(x² + y²) by (3a+2b)
Solution:
(x² + y²) x (3a + 2b)
= x² (3a + 2b) + y² (3a + 2b)
= 3x²a + 2x²b + 3y²a + 2y²b
3ax² + 3av² + 2bx² + 2by²

Question 9.
[-3d + (-7ƒ)] by (5d +ƒ)
Solution:
[-3d + (-7ƒ)] x (5d +ƒ)
= -3d x (5d +ƒ) + (-7ƒ) x (5d +ƒ)
= -15d²-3dƒ- 35dƒ- 7ƒ²
= -15d² – 38dƒ- 7ƒ²

Question 10.
(0.8a – 0.5b) by (1.5a -3b)
Solution:
(0.8a – 0.5b) x (1.5a-3b)
= 0.8a x (1.5a – 36) – 0.56 (1.5a -3b)
= 1.2a² – 2.4ab – 0.75ab + 1.5b²
= 1.2a²-3.15ab+ 1.5b²

Question 11.
(2x² y² – 5xy²) by (x² -y²)
Solution:
(2x² y² – 5xy²) x (x² -y²)
= 2x²y² (x² – y²) – 5x_y² (x² – y²)
= 2x²y² x x² – 2x²y² xy²– 5xy² x x² + 5x² xy²
= 2x^{2 + 2} y²– 2x² x y^{2 + 2}– 5x¹⁺² y²+5xy^{2 + 2}
= 2x⁴y²– 2x²y⁴ – 5x³y²+ 5xy⁴

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Question 15.
(2x²-1) by (4x³ + 5x²)
Solution:
(2x²-1)x(4x³ + 5x²)
= 2x² x (4x³ + 5x²) – 1 (4x³ + 5x²)
= 2x² x 4x³ + 2x² x 5x² – 4x³ – 5x²
= 8x^{2 + 3} + 10x^{2 + 2}-4x³-5x²
= 8x⁵ + 10x⁴ – 4x³ – 5x²

Question 16.
(2xy + 3y²) (3y² – 2)
Solution:
(2xy + 3y²) (3y² – 2)
= 2xy x (3y²-2) + 3y² x (3y²-2)
= 2xy x Zy²+ 2xy x (-2) + Zy² x Zy² – Zy² x 2
= 6xy^{1 + 2}– 4xy + 9y^{2 + 2}– 6y²
= 6xy³ – 4xy + 9y⁴– 6y²
Find the following products and verify the result for x = -1, y = -2 :

Question 17.
(3x-5y)(x+y)
Solution:
(3x-5y)(x+y)
= 3x x (x + y) – 5y x (x + y)
= 3x x x + 3x x y-5y x x-5y x y
= 3x² + 3xy – 5xy – 5y²
= 3x² – 2xy – 5y²
Verfification:
x = -1,y = -2
L.H.S. = (3x-5y)(x+y)
= [3 (-1) -5 (-2)] [-1 – 2]
= (-3 + 10) (-3) = 7 x (-3) = -21
R.H.S. = 3x² – 2xy – 5y²
= 3 (-1)² – 2 (-1) (-2) -5 (-2)²
=3×1-4-5×4=3-4-20
= 3-24 = -21
∴ L.H.S. = R.H.S.

Question 18.
(x²y-1) (3-2x²y)
Solution:
(x²y-1) (3-2x²y)
= x²y (3 – 2x²y) -1(3-2x²y)
= x²y x 3 – x²y x 2x²y – 1 x 3 + 1 x 2x²y
= 3x²y-2x^{2 + 2}x y^1 +1-3 + 2x²y
= 3x²y – 2x⁴y²– 3 + 2x²y
= 3x²y + 2x²y – 2x⁴y² – 3
= 5x²y – 2x⁴y² – 3
Verification : (x = -1, y = -2)
L.H.S. = (x²y – 1) (3 – 2x²y)
= [(-1)² x (-2) -1] [3 – 2 x (-1)2 x (-2)]
= [1 x (-2) -1) [3 – 2 x 1 x (-2)]
= (-2 – 1) (3 + 4) = -3 x 7 = -21
R.H.S. = 5x²y – 2x⁴y² – 3
= 5 (-1)² (-2) -2 (-1)⁴ (-2)² -3
5 x 1 (-2) – 2 (1 x 4) -3
= -10-8-3 = -21
∴ L.H.S. = R.H.S

Question 19.

Solution:

Simplify :

Question 20.
x² (x + 2y) (x – 3y)
Solution:
x² (x + 2y) (x – 3y)
= x² [x (x – 3y) + 2y (x – 3y)]
= x² [x² – 3xy + 2xy – 6y²]
= x² [x² – xy – 6y²)
= x² x x² – x² x xy – x²6y²
= x⁴ – x³y – 6x²y²

Question 21.
(x² – 2y²) (x + 4y)
Solution:
(x² – 2y²) (x + 4y) x²y²
= [x² (x + 4y) -2y² (x + 4y)] x²y²
= (x³ + 4x²y – 2xy² – 8y³) x²y²
= x²y² x x³ + x²y² x 4x²y – 2x²y² x xy² – 8x²y² x y³
= x^{2 +3} y² + 4x^{2 + 2} y^{2 +1} – 2x^{2 +1} y^{2+ 2} – 8x²y²⁺³
= xy + 4⁴xy³ – 2x³y⁴ – 8x²y⁵

Question 22.
a²b² (a + 2b) (3a + b)
Solution:
a²b² (a + 2b) (3a + b)
= a²b² [a (3a + b).+ 2b (3a + b)]
= a²b² [3a² + ab + 6ab + 2b²]
= a²b² [3a² + lab + 2b²]
= a²b² x 3a² + a²b² x 7ab + a²b² x 2b²
= 3a^{2 + 2}b² + 7a²⁺¹ b²⁺¹+ 2a²b^{2 + 2}
= 3a⁴b² + 7a³b³ + 2a²b⁴

Question 23.
x² (x-y) y² (x + 2y)
Solution:
x² (x -y) y² (x + 2y)
= [x² x x – x² x y] [y² x x + y² x 2y]
= (x³ – x²y) (xy² + 2y³)
= x³ (xy² + 2y³) – x²y (xy² + 2y³)
= x³ x xy² + x³ x 2y³ – x²y x xy² – x²y x 2y³
= x^{3 +1} y² + 2x³y³ – x^{2 +1} y^{1+ 2} – 2x²y^{1 + 3}
= x⁴y² + 2x³y³ – x³y³ – 2x²y⁴
= x⁴y² + x³y³ – 2x²y⁴

Question 24.
(x³ – 2x² + 5x-7) (2x-3)
Solution:
(x³ – 2x² + 5x – 7) (2x – 3)
= (2x – 3) (x³ – 2x² + 5x – 7)
= 2x (x³ – 2x² + 5x – 7) -3 (x³ – 2x² + 5x – 7)
= 2x x x³ – 2x x 2x² + 2x x 5x – 2x x 7 -3 x x³ – 3 x (-2x²) – 3 x 5x – 3 x (-7)
= 2x⁴-4x³ + 10x²– 14x-3x³ + 6x²– 15x + 21
= 2x⁴ – 4x³ – 3x³ + 10x² + 6x²– 14x- 15x + 21
= 2x⁴-7x³ + 16x²-29x+ 21

Question 25.
(5x + 3) (x – 1) (3x – 2)
Solution:
(5x + 3) (x – 1) (3x – 2)
= (5x + 3) [x (3x – 2) -1 (3x – 2)]
= (5x + 3) [3x² – 2x – 3x + 2]
= (5x + 3) [3x² – 5x + 2]
= 5x (3x² – 5x + 2) + 3 (3x² – 5x + 2)
= (5x x 3x² – 5x x Sx + 5x x 2)+ [3 x 3x² + 3 x (-5x) + 3×2]
= 15x³ – 25x² + 10x + 9x² – 15x + 6
= 15x³ – 25x² + 9x² + 10x – 15x + 6
= 15x³ – 16x² – 5x + 6

Question 26.
(5-x) (6-5x) (2-x)
Solution:
(5-x) (6-5x) (2-x)
= [5 (6 – 5x) -x (6 – 5x)] (2 – x)
= [30 – 2$x – 6x + 5x²] (2 – x)
= (30 – 3 1x + 5x²) (2-x)
= 2 (30 – 31x + 5x²) – x (30 – 31x + 5x²)
= 60 – 62x + 10x² – 30x + 3 1x² – 5x³
= 60 – 62x – 30x + 10x² + 3 1x² – 5x³
= 60 – 92x + 41x² – 5x³

Question 27.
(2x² + 3x – 5) (3x² – 5x + 4)
Solution:
(2x² + 3x – 5) (3x² – 5x + 4)
= 2x² (3x² – 5x + 4) + 3x (3x² – 5x + 4) -5 (3x² – 5x + 4)
= 2x² x 3x² – 2x² x 5x + 2x² x 4 + 3x x 3x² – 3x x 5x + 3x x 4 – 5 x 3x² – 5 (-5x) -5×4
= 6x⁴ – 10x³ + 8x² + 9x³ – 15x² + 12x – 15x² + 25x-20
= 6x⁴ – 10x³ + 9x³ + 8x² – 15x² – 15x² + 12x + 25x – 20
= 6x⁴ – x³ – 22x² + 37x – 20

Question 28.
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
Solution:
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
= 3x (2x – 3) -2 (2x – 3) + 5x (x + 1) – 3 (x + 1)
= 6x² – 9x – 4x + 6 + 5x² + 5x – 3x – 3
= 6x² + 5x² – 9x – 4x + 5x – 3x + 6 – 3
= 11x²– 11x + 3

Question 29.
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
Solution:
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
= [5x (x + 2) -3 (x + 2)] – [2x (4x – 3) + 5 (4x – 3)]
= [5x² + 1 0x – 3x – 6] – [8x² – 6x + 20x -15]
= (5x² + 7x – 6) – (8x² + 14x – 15)
= 5x² + lx – 6 – 8x² – 14x + 15
= 5x² – 8x² + 7x – 14x – 6 + 15
= -3x² – 7x + 9

Question 30.
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
Solution:
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
= [3x (4x + 3y) + 2y (4x + 3y)]-[2x (7_x-3y)-y(7_x-3y)]
= (12x² + 9xy + 8xy + 6y²) – (14x² – 6xy – 7xy + 3y²)
= (12x² + 17xy + 6y²) – (14x² – 13xy + 3y²)
= 12x² + 17xy + 6y² – 14x² + 13xy – 3y²
= 12x² – 14x² + 17xy + 13xy + 6y² – 3y²
= -2x² + 30xy + 3y²
= -2x² + 3y² + 30xy

Question 31.
(x²-3x + 2) (5x- 2) – (3x² + 4x-5) (2x- 1)
Solution:
(x²-3x + 2) (5x- 2) – (3x² + 4x-5) (2x- 1)
= [5x (x² – 3x + 2) -2 (x² – 3x + 2)] – [2x (3x² + 4x – 5) -1 (3x² + 4x – 5)]
= [5x³ – 15x² + 10x – 2x² + 6x – 4] – [6x³ + 8x² – 10x – 3x² – 4x + 5]
= [5x³ – 15x² – 2x² + 10xc + 6x – 4] – [6x³ + 8x² – 3x² – 10x – 4x + 5]
= (5x³ – 17x² + 16x-4) – (6x³ + 5x² – 14x + 5)
= 5x³ – 17x² + 16x – 4 – 6x³ – 5x² + 14x – 5
= 5x³ – 6x³ – 17x² – 5x² + 16x + 14x – 4 – 5
= -x³ – 22x² + 30x – 9

Question 32.
x³ – 2x² + 3x – 4) (x – 1) – (2x – 3) (x² – x + 1)
Solution:
(x³ – 2x² + 3x – 4) (x – 1) – (2x – 3) (x² – x + 1)
= [x (x³ – 2x² + 3x – 4) – 1 (x³ – 2x² + 3x – 4)] – [2x (x² – x + 1) – 3 (x² – x + 1)]
= [x⁴ – 2x³ + 3x² – 4x – x³ + 2x² – 3x + 4] [2x³ – 2x² + 2x – 3x² + 3x – 3]
= (x⁴ – 2x³ – x³ + 3x² + 2x² – 4x – 3x + 4) (2x³ – 2x² – 3x² + 2x + 3x – 3)
= (x⁴ – 3x³ + 5x² – 7x + 4) – (2x³ – 5x² + 5x – 3)
= x⁴ – 3x³ + 5x² – 7x + 4 – 2x³ + 5x² – 5x + 3
= x⁴ – 3x³ – 2x³ + 5x² + 5x² – 7x – 5x + 4 + 3
= x⁴ – 5x³ + 10x² – 12x + 7

Exercise 6.6

Question 1.
Write the following squares of bionomials as trinomials :

Solution:
Using the formulas
(a + b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b²
(i) (a + 2)² = (a)² + 2 x a x 2 + (2)²
{(a + b)² = a² + 2ab + b²}
= a² + 4a + 4
(ii) (8a + 3b)² = (8a)² + 2 x 8a * 3b + (3b)² = 64² + 48ab + 9 b²
(iii) (2m+ 1)² = (2m)² + 2 x 2m x1 + (1)²
= 4m² + 4m + 1

Question 2.
Find the product of the following binomials :


Solution:

Question 3.
Using the formula for squaring a binomial, evaluate the following :
(i) (102)²
(ii) (99)²
(iii) (1001)²
(iv) (999)²
(v) (703)²
Solution:
(i) (102)² = (100 + 2)²
= (100)² + 2 x 100 x 2 + (2)²
{(a + b)² = a² + 2ab + b²}
= 10000 + 400 + 4 = 10404
(ii) (99)² = (100 – 1)²
= (100)² – 2 x 100 X 1 +(1)²
{(a – b)² = a² – 2ab + b²}
= 10000 -200+1
= 10001 -200 =9801
(iii) (1001 )² = (1000 + 1)²
{(a + b)² = a² + 2ab + b²}
= (1000)² + 2 x 1000 x 1 + (1)²
= 1000000 + 2000 + 1 = 1002001
(iv) (999)² = (1000 – 1)²
{(a – b)² = a² – 2ab + b²}
= (1000)² – 2 x 1000 x 1 + (1)²
= 1000000 – 2000 + 1
= 1000001 -2000 = 998001

Question 4.
Simplify the following using the formula:
(a – b) (a + b) = a² – b² :
(i) (82)² (18)²
(ii) (467)² (33)²
(iii) (79)² (69)²
(iv) 197 x 203
(v) 113 x 87
(vi) 95 x 105
(vii) 1.8 x 2.2
(viii) 9.8 x 10.2
Solution:
(i) (82)² – (18)² = (82 + 18) (82 – 18)
{(a + b)(a- b) = a² – b²} = 100 x 64 = 6400
(ii) (467)² – (33)² = (467 + 33) (467 – 33)
= 500 x 434 = 217000
(ii) (79)² – (69)² = (79 + 69) (79 – 69)
148 x 10= 1480
(iv) 197 x 203 = (200 – 3) (200 + 3)
= (200)² – (3)²
= 40000-9 = 39991
(v) 113 x 87 = (100 + 13) (100- 13)
= (100)² – (13)²
= 10000- 169 = 9831
(vi) 95 x 105 = (100 – 5) (100 + 5)
= (100)² – (5)²
= 10000 – 25 = 9975
(vii) 8 x 2.2 = (2.0 – 0.2) (2.0 + 0.2)
= (2.0)² – (0.2)²
= 4.00 – 0.04 = 3.96
(viii)9.8 x 10.2 = (10.0 – 0.2) (10.0 + 0.2)
(10.0)² – (0.2)²
= 100.00 – 0.04 = 99.96

Question 5.
Simplify the following using the identities :

Solution:

Question 6.
Find the value of x, if
(i)  4x = (52)² – (48)²
(ii) 14x = (47)² – (33)²
(iii)  5x = (50)² – (40)²
Solution:
(i) 4x = (52)² – (48)²
⇒ 4x = (52 + 48) (52 – 48)

Question 7.
If x + 1x= 20, find the value of x²+ 1×2

Solution:

Question 8.
If x – 1x = 3, find the values of x² + 1×2 and x⁴ + 1×4

Solution:

Question 9.
If x² – 1×2= 18, find the values of x+ 1x  and x– 1x
Solution:

Question 10.
Ifx+y = 4 and xy = 2, find the value of x²+y².
Solution:
x + y = 4
Squaring on both sides,
(x + y)² = (4)²
⇒ x² +y² + 2xy = 16
⇒ x²+y² + 2 x 2 = 16                       (∵ xy = 2)
⇒ x² + y² + 4 = 16
⇒ x²+y² = 16 – 4= 12           ‘
∴ x²+y² = 12

Question 11.
If x-y = 7 and xy = 9, find the value of x²+y².
Solution:
x-y = 7
Squaring on both sides,
(x-y)² = (7)²
⇒ x²+y²-2xy = 49
⇒ x² + y² – 2 x 9 = 49                    (∵ xy = 9)
⇒ x² +y² – 18 = 49
⇒ x² + y² = 49 + 18 = 67
∴ x²+y² = 67

Question 12.
If 3x + 5y = 11 and xy = 2, find the value of 9x² + 25y²
Solution:
3 x + 5y = 11, xy = 2
Squaring on both sides,
(3x + 5y)² = (11)²
⇒ (3x)² + (5y)² + 2 x 3x x 5y = 121
⇒ 9x² + 25y² + 30 x 7 = 121
⇒ 9x² + 25y²+ 30 x 2 = 121           (∵ xy = 2)
⇒ 9x² + 25y² + 60 = 121
⇒ 9x² + 25y² = 121 – 60 = 61
∴ 9x² + 25y² = 61

Question 13.
Find the values of the following expressions :
(i)16x² + 24x + 9, when X’ = 745
(ii) 64x² + 81y² + 144xy when x = 11 and y = 43
(iii) 81x² + 16y²-72xy, whenx= 23 andy= 34
Solution:

Question 14.
If x + 1x = 9, find the values of x⁴ + 1×4.
Solution:

Question 15.
If x + 1x = 12, find the values of x–  1x.
Solution:

Question 16.
If 2x + 3y = 14 and 2x – 3y = 2, find the value of xy.
Solution:
2x + 3y = 14, 2x – 3y= 2
We know that
(a + b)² – (a – b)² = 4ab
∴ (2x + 3y)² – (2x – 3y)² = 4 x 2x x 3y = 24xy
⇒ (14)² – (2)² = 24xy
⇒ 24xj= 196-4= 192
⇒ xy = 19224 = 8
∴  xy = 8

Question 17.
If x² + y² = 29 and xy = 2, find the value of
(i) x+y
(ii) x-y
(iii) x⁴ +y⁴
Solution:
x² + y² = 29, xy = 2
(i) (x + y)² = x² + y² + 2xy
= 29 + 2×2 = 29+ 4 = 33
∴  x + y= ±√33
(ii) (x – y)² = x² + y² – 2xy
= 29- 2×2 = 29- 4 = 25
∴ x-y= ±√25= ±5
(iii) x² + y² = 29
Squaring on both sides,
(x² + y²)² = (29)²
⇒ (x²)² + (y²)² + 2x²y² = 841
⇒ x⁴ +y + 2 (xy)² = 841
⇒ x⁴ + y + 2 (2)² = 841          (∵ xy = 2)
⇒ x⁴ + y + 2×4 = 841
⇒ x⁴ + y + 8 = 841
⇒ x⁴ + y = 841 – 8 = 833
∴ x⁴ +y = 833

Question 18.
What must be added to each of the following expressions to make it a whole square ?’
(i) 4x² – 12x + 7
(ii) 4x² – 20x + 20
Solution:
(i) 4x² – 12x + 7 = (2x)² – 2x 2x x 3 + 7
In order to complete the square,
we have to add  3² – 7 = 9 – 7 = 2
∴ (2x)² – 2 x 2x x 3 + (3)²
= (2x-3)²
∴ Number to be added = 2
(ii) 4x² – 20x + 20
⇒ (2x)² – 2 x 2x x 5 + 20
In order to complete the square,
we have to add (5)² – 20 = 25 – 20 = 5
∴ (2x)² – 2 x 2x x 5 + (5)²
= (2x – 5)²
∴ Number to be added = 5

Question 19.
Simplify :
(i) (x-y) (x + y) (x² + y²) (x⁴ + y⁴)
(ii) (2x – 1) (2x + 1) (4x² + 1) (16x⁴ + 1)
(iii) (7m – 8m)² + (7m + 8m)²
(iv) (2.5p -5q)² – (1.5p – 2.5q)²
(v) (m² – n²m)² + 2m³n²
Solution:
(i) (x – y) (x + y) (x² + y²) (x⁴ +y)
= (x² – y²) (x² + y) (x⁴ + y⁴)
= [(x²)² – (y²)²] (x⁴+y⁴)
= (x⁴-y⁴) (x⁴+y⁴)
= (x⁴)² – (y⁴)² = x⁸ – y⁸
(ii) (2x – 1) (2x + 1) (4x² + 1) (16x⁴ + 1)
= [(2x)² – (1)²] (4x² + 1) (16x⁴ + 1)
= (4x² – 1) (4x² + 1) (16x⁴ + 1)
= [(4x²)²-(1)²] (16x⁴+ 1)
= (16x⁴-1) (16x⁴+ 1)
= (16x⁴)²– (1)² = 256x⁸ – 1
(iii) (7m – 8m)² + (7m + 8n)²
= (7m)² + (8n)² – 2 x 7m x 8n + (7m)² + (8n)² + 2 x 7m x 8n
= 49m² + 64m² – 112mn + 49m² + 64m² + 112mn
= 98 m² + 128n²
(iv) (2.5p – 1.5q)² – (1.5p – 2.5q)²
= (2.5p)² + (1.5q)² – 2 x 2.5p x 1.5q
= [(1.5p)² + (1.5q)² – 2 x 1.5 p x 2.5q]
= (6.25p² + 2.25q² – 7.5 pq) – (2.25p² + 6.25q²-7.5pq)
= 6.25p² + 2.25q² – 7.5pq – 2.25p² – 6.25q² + 7.5pq
= 6.25p² – 2.25p² + 2.25g² – 6.25q²
= 4.00P² – 4.00_q²
= 4p² – 4q² = 4 (p² – q²)
(v) (m² – n²m)² + 2m³M²
= (m²)² + (n²m)² -2 x m² x n²m + 2;m³m²
= m⁴ + n⁴m² – 2m³n² + 2m³n²
= m⁴ + n⁴m² = m⁴ + m²n⁴

Question 20.
Show that :

Solution:

Exercise 6.7

Question 1.
Find the following products :

Solution:

Question 2.
Evaluate the following :
(i) 102 x 106
(ii) 109 x 107
(iii) 35 x 37
(iv) 53 x 55
(v) 103 x 96
(vi) 34 x 36
(vii) 994 x 1006
Solution:
(i) 102 x 106 = (100 + 2) (100 + 6)
= (100)² + (2 + 6) x 100 + 2 x 6
= 10000 + 800 + 12 = 10812

(ii) 109 x 107 = (100 + 9) (100 + 7)
= (100)² + (9 + 7) x 100 + 9 x 7
=10000 + 1600 + 63 = 11663

(iii) 35 x 37 = (30 + 5) (30 + 7)
= (30)² + (5 + 7) x 30 + 5 x 7
= 900 + 12 x 30 + 35
= 900 + 360 + 35 = 1295

(iv) 53 x 55 = (50 + 3) (50 + 5)
= (50)² + (3 + 5) x 50 + 3 x 5
= 2500 + 8 x 50 + 15
= 2500 + 400+ 15 = 2915

(v)103 x 96 = (100 + 3) (100-4)
= (100)² + (3 – 4) x 100 + 3 x (-4)
= 10000+ (-1) x 100-12
= 10000 – 100 – 12 = 10000 – 112 = 9888

(vi) 34 x 36 = (30 + 4) (30 + 6)
= (30)² + (4 + 6) x 30 + 4 x 6
= 900 + 10 x 30 + 24
= 900 + 300 + 24 = 1224

(vii) 994 x 1006 = (1000 – 6) (1000 + 6)
= (1000)² + (-6 + 6) x 1000 + (-6) x 6
= 1000000 + 0-36
= 1000000-36 = 999964