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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