Table of Contents
Exercise MCQ
Question 1
If (x + 1) is a factor of the polynomial (2x2 + kx) then the value of k is
(a) -2
(b) -3
(c) 2
(d) 3Solution 1
Correct option: (c)
Let p(x) = 2x2 + kx
Since (x + 1) is a factor of p(x),
P(-1) = 0
⇒ 2(-1)2 + k(-1) = 0
⇒ 2 – k = 0
⇒ k = 2Question 2
The value of (249)2 – (248)2 is
(a) 12
(b) 477
(c) 487
(d) 497Solution 2
Correct option: (d)
(249)2 – (248)2
= (249 + 248)(249 – 248)
= 497 × 1
= 497Question 3
Solution 3
Question 4
If a + b + c = 0, then a3 + b3 + c3= ?
(a) 0
(b) abc
(c) 2abc
(d)3abcSolution 4
Question 5
(a) 0
(b)
(c)
(d) Solution 5
Correct option: (c)
Question 6
The coefficient of x in the expansion of (x + 3)3 is
(a) 1
(b) 9
(c) 18
(d)27Solution 6
Question 7
Which of the following is a factor of (x + y)3 – (x3 + y3)?
(a) x2 + y2 + 2xy
(b) x2 + y2 – xy
(c) xy2
(d) 3xySolution 7
Correct option: (d)
(x + y)3 – (x3 + y3)
= (x + y)3 – [(x + y)(x2 – xy + y2)]
= (x + y)[(x + y)2 – (x2 – xy + y2)]
= (x + y)[x2 + y2 + 2xy – x2 + xy – y2]
= (x + y)(3xy) Question 8
One of the factors of (25x2 – 1) + (1 + 5x)2 is
(a) 5 +x
(b) 5 – x
(c) 5x – 1
(d) 10xSolution 8
Correct option: (d)
(25x2 – 1) + (1 + 5x)2
= [(5x)2 – (1)2] + (1 + 5x)2
= (5x – 1)(5x + 1) + (1 + 5x)2
= (1 + 5x)[(5x – 1) + (1 + 5x)]
= (1 + 5x)(5x – 1 + 1 + 5x)
= (1 + 5x)(10x)Question 9
If (x + 5) is a factor of p(x) = x3 – 20x + 5k then k = ?
(a) -5
(b) 5
(c) 3
(d) -3Solution 9
Correct option: (b)
Since (x + 5) is a factor of p(x) = x3 – 20x + 5k,
p(-5) = 0
⇒ (-5)3 – 20(-5) + 5k = 0
⇒ -125 + 100 + 5k = 0
⇒ -25 + 5k = 0
⇒ 5k = 25
⇒ k = 5 Question 10
If (x + 2) and (x – 1) are factors of (x3 + 10x2 + mx + n), then
(a) m = 5, n = -3
(b) m = 7, n = -18
(c) m = 17, n = -8
(d)m = 23, n = -19Solution 10
Question 11
104 ⨯ 96 =?
(a) 9894
(b) 9984
(c) 9684
(d)9884Solution 11
Question 12
305 ⨯ 308 = ?
(a) 94940
(b) 93840
(c) 93940
(d)94840Solution 12
Question 13
207 ⨯ 193 = ?
(a) 39851
(b) 39951
(c) 39961
(d)38951Solution 13
Question 14
4a2 + b2 + 4ab + 8a + 4b + 4 =?
(a) (2a + b + 2)2
(b) (2a – b + 2)2
(c) (a + 2b + 2)2
(d)None of theseSolution 14
Question 15
(x2 – 4x – 21) =?
(a) (x – 7)(x – 3)
(b) (x + 7)(x – 3)
(c) (x – 7)(x + 3)
(d)none of theseSolution 15
Question 16
(4x2 + 4x -3) = ?
(a) (2x – 1)(2x – 3)
(b) (2x + 1)(2x – 3)
(c) (2x + 3)(2x – 1)
(d)none of theseSolution 16
Question 17
6x2 + 17x + 5 =?
(a) (2x +3)(3x + 5)
(b) (2x + 5)(3x + 1)
(c) (6x + 5)(x + 1)
(d)none of theseSolution 17
Question 18
(x + 1) is a factor of the polynomial
(a) x3 – 2x2 + x + 2
(b) x3 + 2x2 + x – 2
(c) x3 + 2x2 – x – 2
(d)x3 + 2x2 – x + 2Solution 18
Question 19
3x3 + 2x2 + 3x + 2 =?
(a) (3x – 2)(x2 – 1)
(b) (3x – 2)(x2 + 1)
(c) (3x + 2)(x2 – 1)
(d)(3x + 2)(x2 + 1)Solution 19
Question 20
(a) 1
(b) 0
(c) -1
(d)3Solution 20
Question 21
If x + y + z =9 and xy + yz + zx = 23, then the value of (x3 + y3 + z3 – 3xyz) = ?
(a) 108
(b) 207
(c) 669
(d)729Solution 21
Question 22
then (a3 – b3) = ?
(a) -3
(b) -2
(c) -1
(d) 0Solution 22
Correct option: (d)
Exercise Ex. 3C
Question 1
Factorise:
x2 + 11x + 30Solution 1
x2 + 11x + 30
= x2 + 6x + 5x + 30
= x (x + 6) + 5 (x + 6)
= (x + 6) (x + 5).Question 2
Factorise:
x2 + 18x + 32Solution 2
x2 + 18x + 32
= x2 + 16x + 2x + 32
= x (x + 16) + 2 (x + 16)
= (x + 16) (x + 2).Question 3
Factorise:
x2 + 20x – 69Solution 3
x2 + 20x – 69
= x2 + 23x – 3x – 69
= x(x + 23) – 3(x + 23)
= (x + 23)(x – 3) Question 4
Factorise:
x2 + 19x – 150Solution 4
x2 + 19x – 150
= x2 + 25x – 6x – 150
= x(x + 25) – 6(x + 25)
= (x + 25)(x – 6) Question 5
Factorise:
x2 + 7x – 98Solution 5
x2 + 7x – 98
= x2 + 14x – 7x – 98
= x(x + 14) – 7(x + 14)
= (x + 14)(x – 7) Question 6
Factorise:
Solution 6
Question 7
Factorise:
x2 – 21x + 90Solution 7
x2 – 21x + 90
= x2 – 6x – 15x + 90
= x(x – 6) – 15(x – 6)
= (x – 6)(x – 15) Question 8
Factorise:
x2 – 22x + 120Solution 8
x2 – 22x + 120
= x2 – 10x – 12x + 120
= x(x – 10) – 12(x – 10)
= (x – 10)(x – 12) Question 9
Factorise:
x2 – 4x + 3Solution 9
x2 – 4x + 3
= x2 – 3x – x + 3
= x(x – 3) – 1(x – 3)
= (x – 3)(x – 1) Question 10
Factorise:
Solution 10
Question 11
Factorise:
Solution 11
Question 12
Factorise:
Solution 12
Question 13
Factorise:
Solution 13
Question 14
Factorise:
x2 – 24x – 180Solution 14
x2 – 24x – 180
= x2 – 30x + 6x – 180
= x(x – 30) + 6(x – 30)
= (x – 30)(x + 6) Question 15
Factorise:
z2 – 32z – 105Solution 15
z2 – 32z – 105
= z2 – 35z + 3z – 105
= z (z – 35) + 3 (z – 35)
= (z – 35) (z + 3)Question 16
Factorise:
x2 – 11x – 80Solution 16
x2 – 11x – 80
= x2 – 16x + 5x – 80
= x (x – 16) + 5 (x – 16)
= (x – 16) (x + 5).Question 17
Factorise:
6 – x – x2Solution 17
6 – x – x2
= 6 + 2x – 3x – x2
= 2(3 + x) – x (3 + x)
= (3 + x) (2 – x).Question 18
Factorise:
Solution 18
Question 19
Factorise:
40 + 3x – x2Solution 19
40 + 3x – x2
= 40 + 8x – 5x – x2
= 8 (5 + x) -x (5 + x)
= (5 + x) (8 – x).Question 20
Factorise:
x2 – 26x + 133Solution 20
x2 – 26x + 133
= x2 – 19x – 7x + 133
= x(x – 19) – 7(x – 19)
= (x – 19)(x – 7) Question 21
Factorise:
Solution 21
Question 22
Factorise:
Solution 22
Question 23
Factorise:
Solution 23
Question 24
Factorise:
Solution 24
Question 25
Factorise:
x2 – x – 156Solution 25
x2 – x – 156
= x2 – 13x + 12x – 156
= x (x – 13) + 12 (x – 13)
= (x – 13) (x + 12).Question 27
Factorise:
9x2 + 18x + 8Solution 27
9x2 + 18x + 8
= 9x2 + 12x + 6x + 8
= 3x (3x+ 4) +2 (3x + 4)
= (3x + 4) (3x + 2).Question 28
Factorise:
6x2 + 17x + 12Solution 28
6x2 + 17x + 12
= 6x2 + 9x + 8x + 12
= 3x (2x + 3) + 4(2x + 3)
= (2x + 3) (3x + 4).Question 29
Factorise:
18x2 + 3x – 10Solution 29
18x2 + 3x – 10
= 18x2 – 12x + 15x – 10
= 6x (3x – 2) + 5 (3x – 2)
= (6x + 5) (3x – 2).Question 30
Factorise:
2x2 + 11x – 21Solution 30
2x2 + 11x – 21
= 2x2 + 14x – 3x – 21
= 2x (x + 7) – 3 (x + 7)
= (x + 7) (2x – 3).Question 31
Factorise:
15x2 + 2x – 8Solution 31
15x2 + 2x – 8
= 15x2 – 10x + 12x – 8
= 5x (3x – 2) + 4 (3x – 2)
= (3x – 2) (5x + 4).Question 32
Factorise:
21x2 + 5x – 6Solution 32
21x2 + 5x – 6
= 21x2 + 14x – 9x – 6
= 7x(3x + 2) – 3(3x + 2)
= (3x + 2)(7x – 3) Question 33
Factorise:
24x2 – 41x + 12Solution 33
24x2 – 41x + 12
= 24x2 – 32x – 9x + 12
= 8x (3x – 4) – 3 (3x – 4)
= (3x – 4) (8x – 3).Question 34
Factorise:
3x2 – 14x + 8Solution 34
3x2 – 14x + 8
= 3x2 – 12x – 2x +8
= 3x (x – 4) – 2(x – 4)
= (x – 4) (3x – 2).Question 35
Factorise:
2x2 + 3x – 90Solution 35
2x2 + 3x – 90
= 2x2 – 12x + 15x – 90
= 2x (x – 6) + 15 (x – 6)
= (x – 6) (2x + 15).Question 37
Factorise:
Solution 37
Question 38
Factorise:
Solution 38
Question 39
Factorise:
Solution 39
Question 40
Factorise:
Solution 40
Question 41
Factorise:
Solution 41
Question 42
Factorise:
Solution 42
Question 43
Factorise:
Solution 43
Question 44
Factorise:
15x2 – x – 28Solution 44
15x2 – x – 28
= 15x2 + 20x – 21x – 28
= 5x (3x + 4) – 7 (3x + 4)
= (3x + 4) (5x – 7).Question 45
Factorise:
6x2 – 5x – 21Solution 45
6x2 – 5x – 21
= 6x2 + 9x – 14x – 21
= 3x (2x + 3) – 7 (2x + 3)
= (3x – 7) (2x + 3).Question 46
Factorise:
2x2 – 7x – 15Solution 46
2x2 – 7x – 15
= 2x2 – 10x + 3x – 15
= 2x (x – 5) + 3 (x – 5)
= (x – 5) (2x + 3).Question 47
Factorise:
5x2 – 16x – 21Solution 47
5x2 – 16x – 21
= 5x2 + 5x – 21x – 21
= 5x (x + 1) -21 (x + 1)
= (x + 1) (5x – 21).Question 48
Factorise:
6x2 – 11x – 35Solution 48
6x2 – 11x – 35
= 6x2 – 21x + 10x – 35
= 3x(2x – 7) + 5(2x – 7)
= (2x – 7)(3x + 5) Question 49
Factorise:
9x2 – 3x – 20Solution 49
9x2 – 3x – 20
= 9x2 – 15x + 12x – 20
= 3x(3x – 5) + 4(3x – 5)
= (3x – 5)(3x + 4) Question 50
Factorise:
10x2 – 9x – 7Solution 50
10x2 – 9x – 7
= 10x2 + 5x – 14x – 7
= 5x (2x + 1) – 7 (2x+ 1)
= (2x + 1) (5x – 7).Question 51
Factorise:
x2 – 2x + Solution 51
x2 – 2x +
Question 52
Factorise:
Solution 52
Question 53
Factorise:
Solution 53
Question 54
Factorise:
Solution 54
Question 55
Factorise:
Solution 55
Question 56
Factorise:
Solution 56
Question 57
Factorise:
Solution 57
Question 58
Factorise:
Solution 58
Question 59
Factorise:
Solution 59
Letx + y = z
Then, 2 (x+ y)2 – 9 (x + y) – 5
Now, replacing z by (x + y), we get
Question 60
Factorise:
9 (2a – b)2 – 4 (2a – b) – 13Solution 60
Let2a – b = c
Then,9 (2a – b)2 – 4 (2a – b) -13
Now, replacing c by (2a – b) , we get
9 (2a – b)2 – 4 (2a – b) – 13
Question 61
Factorise:
7 (x – 2y)2 – 25 (x – 2y) + 12Solution 61
Let x – 2y = z
Then, 7 (x – 2y)2 – 25 (x – 2y) + 12
Now replace z by (x – 2y), we get
7 (x – 2y)2 – 25 (x – 2y) + 12
Question 62
Factorise:
Solution 62
Question 63
Factorise:
Solution 63
Question 64
Factorise:
(a + 2b)2 + 101(a + 2b) + 100Solution 64
Given equation: (a + 2b)2 + 101(a + 2b) + 100
Let (a + 2b) = x
Then, we have
x2 + 101x + 100
= x2 + 100x + x + 100
= x(x + 100) + 1(x + 100)
= (x + 100)(x + 1)
= (a + 2b + 100)(a + 2b + 1) Question 65
Factorise:
Solution 65
Let x2 = y
Then, 4x4 + 7x2 – 2
Now replacing y by x2, we get
Question 66
Evaluate {(999)2 – 1}.Solution 66
{(999)2 – 1}
= {(999)2 – (1)2}
= {(999 – 1)(999 + 1)}
= 998 × 1000
= 998000
Exercise Ex. 3D
Question 1(i)
Expand:
(a + 2b + 5c)2Solution 1(i)
We know:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (a + 2b + 5c)2
= (a)2 + (2b)2 + (5c)2 + 2(a) (2b) + 2 (2b) (5c) + 2(5c) (a)
= a2 + 4b2 + 25c2 + 4ab + 20bc + 10acQuestion 1(ii)
Expand:
(2a – b + c)2Solution 1(ii)
We know:
(2a – b + c)2
= (2a)2 + (-b)2 + (c)2 + 2 (2a) (-b) + 2(-b) (c) + 2 (c) (2a)
= 4a2 + b2 + c2 – 4ab – 2bc + 4ac.Question 1(iii)
Expand:
(a – 2b – 3c)2Solution 1(iii)
We know:
(a – 2b – 3c)2
= (a)2 + (-2b)2 + (-3c)2+ 2(a) (-2b) + 2(-2b) (-3c) + 2 (-3c) (a)
= a2 + 4b2 + 9c2 – 4ab + 12bc – 6ac.Question 2(i)
Expand:
(2a – 5b – 7c)2Solution 2(i)
We know:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(2a – 5b – 7c)2
= (2a)2 + (-5b)2 + (-7c)2 + 2 (2a) (-5b) + 2 (-5b) (-7c) + 2 (-7c) (2a)
= 4a2 + 25b2 + 49c2 – 20ab + 70bc – 28ac.Question 2(ii)
Expand:
(ii) (-3a + 4b – 5c)2Solution 2(ii)
(-3a + 4b – 5c)2
= (-3a)2 + (4b)2 + (-5c)2 + 2 (-3a) (4b) + 2 (4b) (-5c) + 2 (-5c) (-3a)
= 9a2 + 16b2 + 25c2 – 24ab – 40bc + 30ac.Question 2(iii)
Expand:
Solution 2(iii)
= Question 3
Factorise: 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz.Solution 3
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2 (2x) (3y) + 2(3y) (-4z) + 2 (-4z) (2x)
= (2x + 3y – 4z)2Question 4
Factorise: 9x2 + 16y2 + 4z2 – 24xy + 16yz – 12xzSolution 4
9x2 + 16y2 + 4z2 – 24xy + 16yz – 12xz
= (-3x)2 + (4y)2 + (2z)2 + 2 (-3x) (4y) + 2 (4y) (2z) + 2 (2z) (-3x)
= (-3x + 4y + 2z)2.Question 5
Factorise: 25x2 + 4y2 + 9z2 – 20xy – 12yz + 30xz.Solution 5
25x2 + 4y2 + 9z2 – 20xy – 12yz + 30xz
= (5x)2 + (-2y)2 + (3z)2 + 2(5x) (-2y) + 2(-2y) (3z) + 2(3z) (5x)
= (5x – 2y + 3z)2Question 6
16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xzSolution 6
16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz
= (4x)2 + (-2y)2 + (3z)2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(4x)(3z)
= [4x + (-2y) + 3z]2
= (4x – 2y + 3z)2
= (4x – 2y + 3z)(4x – 2y + 3z) Question 7(i)
Evaluate
(99)2Solution 7(i)
(99)2
= (100 – 1)2
= (100)2 – 2(100) (1) + (1)2
= 10000 – 200 + 1
= 9801.Question 7(ii)
Evaluate
(995)2Solution 7(ii)
(995)2
= (1000 – 5)2
= (1000)2 + (5)2 – 2(1000)(5)
= 1000000 + 25 – 10000
= 990025 Question 7(iii)
Evaluate
(107)2Solution 7(iii)
(107)2
= (100 + 7)2
= (100)2 + (7)2 + 2(100)(7)
= 10000 + 49 + 1400
= 11449
Exercise Ex. 3E
Question 1(i)
Expand:
(3x + 2)3Solution 1(i)
(3x + 2)3
= (3x)3 + (2)3 + 3 3x 2 (3x + 2)
= 27x3 + 8 + 18x (3x + 2)
= 27x3 + 8 + 54x2 + 36x.Question 1(ii)
Expand:
Solution 1(ii)
Question 1(iii)
Expand
Solution 1(iii)
Question 2(i)
Expand
(5a – 3b)3Solution 2(i)
(5a – 3b)3
= (5a)3 – (3b)3 – 3(5a)(3b)(5a – 3b)
= 125a3 – 27b3 – 45ab(5a – 3b)
= 125a3 – 27b3 – 225a2b + 135ab2 Question 2(ii)
Expand
Solution 2(ii)
Question 2(iii)
Expand
Solution 2(iii)
Question 3
Factorise
8a3 + 27b3 + 36a2b + 54ab2Solution 3
8a3 + 27b3 + 36a2b + 54ab2
= (2a)3 + (3b)3 + 3 × (2a)2 × (3b) + 3 × (2a) × (3b)2
= (2a + 3b)3
= (2a + 3b)(2a + 3b)(2a + 3b) Question 4
Factorise
64a3 – 27b3 – 144a2b + 108ab2Solution 4
64a3 – 27b3 – 144a2b + 108ab2
= (4a)3 – (3b)3 – 3 × (4a)2 × (3b) + 3 × (4a) × (3b)2
= (4a – 3b)3
= (4a – 3b)(4a – 3b)(4a – 3b) Question 5
Factorise
Solution 5
Question 6
Factorise
125x3 – 27y3 – 225x2y + 135xy2Solution 6
125x3 – 27y3 – 225x2y + 135xy2
= (5x)3 – (3y)3 – 3 × (5x)2 × (3y) + 3 × (5x) × (3y)2
= (5x – 3y)3
= (5x – 3y)(5x – 3y)(5x – 3y) Question 7
Factorise
a3x3 – 3a2bx2 + 3ab2x – b3Solution 7
a3x3 – 3a2bx2 + 3ab2x – b3
= (ax)3 – 3 × (ax)2 × b + 3 × (ax) × (b)2 – (b)3
= (ax – b)3
= (ax – b)(ax – b)(ax – b) Question 8
Factorise
Solution 8
Question 9
Factorise
a3 – 12a(a – 4) – 64Solution 9
a3 – 12a(a – 4) – 64
= a3 – 3 × a × 4(a – 4) – (4)3
= (a – 4)3
= (a – 4)(a – 4)(a – 4) Question 10(i)
Evaluate
(103)3Solution 10(i)
(103)3
= (100 + 3)3
= (100)3 + (3)3 + 3 × 100 × 3 × (100 + 3)
= 1000000 + 27 + 900(103)
= 1000000 + 27 + 92700
= 1092727 Question 10(ii)
Evaluate
(99)3Solution 10(ii)
(99)3
= (100 – 1)3
= (100)3 – (1)3 – 3 × 100 × 1 × (100 – 1)
= 1000000 – 1 – 300(99)
= 1000000 – 1 – 29700
= 970299
Exercise Ex. 3F
Question 1
Factorise:
x3 + 27Solution 1
x3 + 27
= x3 + 33
= (x + 3) (x2 – 3x + 9)
Question 2
Factorise
27a3 + 64b3Solution 2
27a3 + 64b3
= (3a)3 + (4b)3
= (3a + 4b)[(3a)2 – (3a)(4b) + (4b)2]
= (3a + 4b)(9a2 – 12ab + 16b2) Question 3
Factorise:
125a3 + Solution 3
125a3 +
We know that
Let us rewrite
Question 4
Factorise:
216x3 + Solution 4
216x3 +
We know that
Let us rewrite
Question 5
Factorise:
16x4 + 54xSolution 5
16x 4 + 54x
= 2x (8x 3 + 27)
= 2x [(2x)3 + (3)3]
= 2x (2x + 3) [(2x)2 – 2x3 + 32]
=2x(2x+3)(4x2 -6x +9)
Question 6
Factorise:
7a3 + 56b3Solution 6
7a3 + 56b3
= 7(a3 + 8b3)
= 7 [(a)3 + (2b)3]
= 7 (a + 2b) [a2 – a 2b + (2b)2]
= 7 (a + 2b) (a2 – 2ab + 4b2).Question 7
Factorise:
x5 + x2Solution 7
x5 + x2
= x2(x3 + 1)
= x2 (x + 1) [(x)2 – x 1 + (1)2]
= x2 (x + 1) (x2 – x + 1).Question 8
Factorise:
a3 + 0.008Solution 8
a3 + 0.008
= (a)3 + (0.2)3
= (a + 0.2) [(a)2 – a 0.2 + (0.2)2]
= (a + 0.2) (a2 – 0.2a + 0.04).Question 9
Factorise:
1 – 27x3Solution 9
1 – 27x3
= (1)3 – (3x)3
= (1 – 3x) [(1)2 + 1 3x + (3x)2]
= (1 – 3x) (1 + 3x + 9x2).Question 10
Factorise
64a3 – 343Solution 10
64a3 – 343
= (4a)3 – (7)3
= (4a – 7)[(4a)2 + (4a)(7) + (7)2]
= (4a – 7)(16a2 + 28a + 49) Question 11
Factorise:
x3 – 512Solution 11
x3 – 512
= (x)3 – (8)3
= (x – 8) [(x)2 + x 8 + (8)2]
= (x – 8) (x2 + 8x + 64).Question 12
Factorise:
a3 – 0.064Solution 12
a3 – 0.064
= (a)3 – (0.4)3
= (a- 0.4) [(a)2 + a 0.4 + (0.4)2]
= (a – 0.4) (a2 + 0.4 a + 0.16).Question 13
Factorise:
Solution 13
We know that
Let us rewrite
Question 14
Factorise
Solution 14
Question 15
Factorise:
x – 8xy3Solution 15
x – 8xy3
= x (1 – 8y3)
= x [(1)3 – (2y)3]
= x (1 – 2y) [(1)2 + 1 2y + (2y)2]
= x (1 – 2y) (1 + 2y + 4y2).Question 16
Factorise:
32x4 – 500xSolution 16
32x4 – 500x
= 4x (8x3 – 125)
= 4x [(2x)3 – (5)3]
= 4x [(2x – 5) [(2x)2 + 2x 5 + (5)2]
= 4x (2x – 5) (4x2 + 10x + 25).Question 17
Factorise:
3a7b – 81a4b4Solution 17
3a7b – 81a4b4
= 3a4b (a3 – 27b3)
= 3a4b [(a)3 – (3b)3]
= 3a4b (a – 3b) [(a)2 + a 3b + (3b)2]
= 3a4b (a – 3b) (a2 + 3ab + 9b2).Question 18
Factorise
x4y4 – xySolution 18
x4y4 – xy
= xy(x3y3 – 1)
= xy[(xy)3 – (1)3]
= xy{(xy – 1)[(xy)2 + (xy)(1) + (1)2]}
= xy(xy – 1)(x2y2 + xy + 1) Question 19
Factorise
8x2y3 – x5Solution 19
8x2y3 – x5
= x2 (8y3 – x3)
= x2 [(2y)3 – x3]
= x2 [(2y – x)[(2y)2 + (2y)(x) + x2]
= x2 (2y – x)(4y2 + 2xy + x2) Question 20
Factorise
1029 – 3x3Solution 20
1029 – 3x3
= 3(343 – x3)
= 3[(7)3 – x3]
= 3[(7 – x)(72 + 7x + x2)]
= 3(7 – x)(49 + 7x + x2) Question 21
Factorise:
x6 – 729Solution 21
x6 – 729
= (x2)3 – (9)3
= (x2 – 9) [(x2)2 + x2 9 + (9)2]
= (x2 – 9) (x4 + 9x2 + 81)
= (x + 3) (x – 3) [(x2 + 9)2 – (3x)2]
= (x + 3) (x – 3) (x2 + 3x + 9) (x2 – 3x + 9).Question 22
Factorise
x9 – y9Solution 22
x9 – y9
= (x3)3 – (y3)3
= [(x3 – y3)][(x3)2 + x3y3 + (y3)2]
= [(x – y)(x2 + xy + y2)(x6 + x3y3 + y6) Question 23
Factorise:
(a + b)3 – (a – b)3Solution 23
We know that,
Therefore,
(a + b)3 – (a – b)3
= [a + b – (a – b)] [ (a + b)2 + (a + b) (a – b) + (a – b)2]
= (a + b – a + b) [ a2 + b2 + 2ab + a2 – b2 + a2 + b2 – 2ab]
= 2b (3a2 + b2).Question 24
Factorise:
8a3 – b3 – 4ax + 2bxSolution 24
8a3 – b3 – 4ax + 2bx
= 8a3 – b3 – 2x (2a – b)
= (2a)3 – (b)3 – 2x (2a – b)
= (2a – b) [(2a)2 + 2a b + (b)2] – 2x (2a – b)
= (2a – b) (4a2 + 2ab + b2) – 2x (2a – b)
= (2a – b) (4a2 + 2ab + b2 – 2x).Question 25
Factorise:
a3 + 3a2b + 3ab2 + b3 – 8Solution 25
Question 26
Factorise:
Solution 26
We know that
Question 27
Factorise:
2a3 + 16b3 – 5a – 10bSolution 27
2a3 + 16b3 – 5a – 10b
= 2 (a3 + 8b3) – 5 (a + 2b)
= 2 [(a)3 + (2b)3] – 5 (a + 2b)
= 2 (a + 2b) [(a)2 – a 2b + (2b)2 ] – 5 (a + 2b)
= (a+ 2b) [2(a2 – 2ab + 4b2) – 5]Question 28
Factorise
a6 + b6Solution 28
a6 + b6
= (a2)3 + (b2)3
= (a2 + b2)[(a2)2 – (a2b2) + (b2)2]
= (a2 + b2)(a4 – a2b2 + b4) Question 29
Factorise
a12 – b12Solution 29
a12 – b12
= (a6)2 – (b6)2
= (a6 – b6)(a6 + b6)
= [(a3)2 – (b3)2][(a2)3 + (b2)3]
= (a3 – b3)(a3 + b3)[(a2 + b2)(a4 – a2b2 + b4)]
= (a – b)(a2 + ab + b2)(a + b)(a2 – ab + b2)(a2 + b2)(a4 – a2b2 + b4)
= (a – b)(a + b)(a2 + b2)(a2 + ab + b2)(a2 – ab + b2)(a4 – a2b2 + b4)Question 30
Factorise
x6 – 7x3 – 8Solution 30
Given equation is x6 – 7x3 – 8.
Putting x3 = y, we get
y2 – 7y – 8
= y2 – 8y + y – 8
= y(y – 8) + 1(y – 8)
= (y – 8)(y + 1)
= (x3 – 8)(x3 + 1)
= (x3 – 23)(x3 + 13)
= (x – 2)(x2 + 2x + 4)(x + 1)(x2 – x + 1)
= (x – 2)(x + 1)(x2 + 2x + 4)(x2 – x + 1) Question 31
Factorise
x3 – 3x2 + 3x + 7Solution 31
x3 – 3x2 + 3x + 7
= x3 – 3x2 + 3x – 1 + 8
= (x3 – 3x2 + 3x – 1) + 8
= (x – 1)3 + 23
= (x – 1 + 2)[(x – 1)2 – (x – 1)(2) + 22]
= (x + 1)(x2 – 2x + 1 – 2x + 2 + 4)
= (x + 1)(x2 – 4x + 7) Question 32
Factorise
(x + 1)3 + (x – 1)3Solution 32
(x + 1)3 + (x – 1)3
= (x + 1 + x – 1)[(x + 1)2 – (x + 1)(x – 1) + (x – 1)2]
= 2x(x2 + 2x + 1 – x2 + 1 + x2 – 2x + 1)
= 2x(x2 + 3) Question 33
Factorise
(2a + 1)3 + (a – 1)3Solution 33
(2a + 1)3 + (a – 1)3
= (2a + 1 + a – 1)[(2a + 1)2 – (2a + 1)(a – 1) + (a – 1)2]
= (3a)[4a2 + 4a + 1 – 2a2 + 2a – a + 1 + a2 – 2a + 1]
= 3a(3a2 + 3a + 3)
= 9a(a2 + a + 1) Question 34
Factorise
8(x + y)3 – 27(x – y)3Solution 34
8(x + y)3 – 27(x – y)3
= [23 (x + y)3] – [33 (x – y)3]
= [2(x + y) – 3(x – y)]{[2(x + y)]2 + 2(x + y)3(x – y) + [3(x – y)]2}
= (2x + 2y – 3x + 3y){[4(x2 + y2 + 2xy)] + 6(x2 – y2) + [9(x2 + y2 – 2xy]}
= (-x + 5y){4x2 + 4y2 + 8xy + 6x2 – 6y2 + 9x2 + 9y2 – 18xy}
= (-x + 5y)(19x2 + 7y2 – 10xy) Question 35
Factorise
(x + 2)3 + (x – 2)3Solution 35
(x + 2)3 + (x – 2)3
= [(x + 2) + (x – 2)][(x + 2)2 – (x + 2)(x – 2) + (x – 2)2]
=(2x)(x2 + 4x + 4 – x2 + 4 + x2 – 4x + 4)
= 2x(x2 + 12) Question 36
Factorise
(x + 2)3 – (x – 2)3Solution 36
(x + 2)3 – (x – 2)3
= [(x + 2) – (x – 2)][(x + 2)2 + (x + 2)(x – 2) + (x – 2)2]
= 4(x2 + 4x + 4 + x2 – 4 + x2 – 4x + 4)
= 4(3x2 + 4) Question 37
Prove that
Solution 37
Question 38
Prove that
Solution 38
Exercise Ex. 3G
Question 1
Find the product:
(x + y – z) (x2 + y2 + z2 – xy + yz + zx)Solution 1
(x + y – z) (x2 + y2 + z2 – xy + yz + zx)
= [x + y + (-z)] [(x)2 + (y)2 + (-z)2 – (x) (y) – (y) (-z) – (-z) (x)]
= x3 + y3 – z3 + 3xyz.Question 2
Find the product.
(x – y – z)(x2 + y2 + z2 + xy – yz + xz)Solution 2
(x – y – z)(x2 + y2 + z2 + xy – yz + xz)
= x3 + xy2 + xz2 + x2y – xyz + x2z – x2y – y3 – yz2 – xy2 + y2z – xyz – x2z – y2z – z3 – xyz + yz2 – xz2
= x3 – y3 – z3 – xyz – xyz – xyz
= x3 – y3 – z3 – 3xyz Question 3
Find the product:
(x – 2y + 3) (x2 + 4y2 + 2xy – 3x + 6y + 9)Solution 3
(x – 2y + 3) (x2 + 4y2 + 2xy – 3x + 6y + 9)
= [x + (-2y) + 3] [(x)2 + (-2y)2 + (3) – (x) (-2y) – (-2y) (3) – (3) (x)]
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
= a3 + b3 + c3 – 3abc
Where, x = a, (-2y) = b and 3 = c
(x – 2y + 3) (x2 + 4y2 + 2xy – 3x + 6y + 9)
= (x)3 + (-2y)3 + (3)2 – 3 (x) (-2y) (3)
= x3 – 8y3 + 27 + 18xy.Question 4
Find the product.
(3x – 5y + 4)(9x2 + 25y2 + 15xy + 20y – 12x + 16)
*Modified the questionSolution 4
(3x – 5y + 4)(9x2 + 25y2 + 15xy + 20y – 12x + 16)
= (3x + (-5y) + 4)[(3x)2 + (-5y)2 + (4)2 – (3x)(-5y) – (-5y)(4) – (3x)(4)]
= (3x)3 + (-5y)3 + (4)3 – 3(3x)(-5y)(4)
= 27x3 – 125y3 + 64 + 180xy Question 5
Factorise:
125a3 + b3 + 64c3 – 60abcSolution 5
125a3 + b3 + 64c3 – 60abc
= (5a)3 + (b)3 + (4c)3 – 3 (5a) (b) (4c)
= (5a + b + 4c) [(5a)2 + b2 + (4c)2 – (5a) (b) – (b) (4c) – (5a) (4c)]
[ a3 + b3 + c3 – 3abc = (a+ b + c) (a2 + b2 + c2 – ab – bc – ca)]
= (5a + b + 4c) (25a2 + b2 + 16c2 – 5ab – 4bc – 20ac).Question 6
Factorise:
a3 + 8b3 + 64c3 – 24abcSolution 6
a3 + 8b3 + 64c3 – 24abc
= (a)3 + (2b)3 + (4c)3 – 3 a 2b 4c
= (a + 2b + 4c) [a2 + 4b2 + 16c2 – 2ab – 8bc – 4ca).Question 7
Factorise:
1 + b3 + 8c3 – 6bcSolution 7
1 + b3 + 8c3 – 6bc
= 1 + (b)3 + (2c)3 – 3 (b) (2c)
= (1 + b + 2c) [1 + b2 + (2c)2 – b – b 2c – 2c]
= (1 + b + 2c) (1 + b2 + 4c2 – b – 2bc – 2c).Question 8
Factorise:
216 + 27b3 + 8c3 – 108bcSolution 8
216 + 27b3 + 8c3 – 108bc
= (6)3 + (3b)3 + (2c)2 – 3 6 3b 2c
= (6 + 3b + 2c) [(6)2 + (3b)2 + (2c)2 – 6 3b – 3b 2c – 2c 6]
= (6 + 3b + 2c) (36 + 9b2 + 4c2 – 18b – 6bc – 12c).Question 9
Factorise:
27a3 – b3 + 8c3 + 18abcSolution 9
27a3 – b3 + 8c3 + 18abc
= (3a)3 + (-b)3 + (2c)3 + 3(3a) (-b) (2c)
= [3a + (-b) + 2c] [(3a)2 + (-b)2 + (2c)2 – 3a (-b) – (-b) (2c) – (2c) (3a)]
= (3a – b + 2c) (9a2 + b2 + 4c2 + 3ab + 2bc – 6ca).Question 10
Factorise:
8a3 + 125b3 – 64c3 + 120abcSolution 10
8a3 + 125b3 – 64c3 + 120abc
= (2a)3 + (5b)3 + (-4c)3 – 3 (2a) (5b) (-4c)
= (2a + 5b – 4c) [(2a)2 + (5b)2 + (-4c)2 – (2a) (5b) – (5b) (-4c) – (-4c) (2a)]
= (2a + 5b – 4c) (4a2 + 25b2 + 16c2 – 10ab + 20bc + 8ca).Question 11
Factorise:
8 – 27b3 – 343c3 – 126bcSolution 11
8 – 27b3 – 343c3 – 126bc
= (2)3 + (-3b)3 + (-7c)3 – 3(2) (-3b) (-7c)
= (2 – 3b – 7c) [(2)2 + (-3b)2 + (-7c)2 – (2) (-3b) – (-3b) (-7c) – (-7c) (2)]
= (2 – 3b – 7c) (4 + 9b2 + 49c2 + 6b – 21bc + 14c).Question 12
Factorise:
125 – 8x3 – 27y3 – 90xySolution 12
125 – 8x3 – 27y3 – 90xy
= (5)3 + (-2x)3 + (-3y)3 – 3 (5) (-2x) (-3y)
= (5 – 2x – 3y) [(5)2 + (-2x)2 + (-3y)2 – (5) (-2x) – (-2x) (-3y) – (-3y) (5)]
= (5 – 2x – 3y) (25 + 4x2 + 9y2 + 10x – 6xy + 15y).Question 13
Factorise:
Solution 13
Question 14
Factorise:
27x3 – y3 – z3 – 9xyzSolution 14
27x3 – y3 – z3 – 9xyz
= (3x)3 – y3 – z3 – 3(3x)(y)(z)
= (3x)3 + (-y)3 + (-z)3 – 3(3x)(-y)(-z)
= [3x + (-y) + (-z)][(3x)2 + (-y)2 + (-z)2 – (3x)(-y) – (-y)(-z) – (3x)(-z)]
= (3x – y – z)(9x2 + y2 + z2 + 3xy – yz + 3zx) Question 15
Factorise:
Solution 15
Question 16
Factorise:
Solution 16
Question 17
Factorise:
(a – b)3 + (b – c)3 + (c – a)3Solution 17
Putting (a – b) = x, (b – c) = y and (c – a) = z, we get,
(a – b)3 + (b – c)3 + (c – a)3
= x3 + y3 + z3, where (x + y + z) = (a – b) + (b – c) + (c – a) = 0
= 3xyz[ (x + y + z)= 0 (x3 + y3 + z3) = 3xyz]
= 3(a – b) (b – c) (c – a).Question 18
Factorise:
(a – 3b)3 + (3b – c)3 + (c – a)3Solution 18
Given equation is (a – 3b)3 + (3b – c)3 + (c – a)3
Now,
(a – 3b) + (3b – c) + (c – a)
= a – a – 3b + 3b – c + c
= 0
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz
Hence,
(a – 3b)3 + (3b – c)3 + (c – a)3
= 3(a – 3b)(3b – c)(c – a)Question 19
Factorise:
(3a – 2b)3 + (2b – 5c)3 + (5c – 3a)3Solution 19
We have:
(3a – 2b) + (2b – 5c) + (5c – 3a) = 0
So, (3a – 2b)3 + (2b – 5c)3 + (5c – 3a)3
= 3(3a – 2b) (2b – 5c) (5c – 3a).Question 20
Factorise:
(5a – 7b)3 + (7b – 9c)3 + (9c – 5a)3Solution 20
Given equation is (5a – 7b)3 + (7b – 9c)3 + (9c – 5a)3
Now,
(5a – 7b) + (7b – 9c) + (9c – 5a)
= 5a – 7b + 7b – 9c + 9c – 5a
= 0
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz
Hence,
(5a – 7b)3 + (7b – 9c)3 + (9c – 5a)3
= 3(5a – 7b)(7b – 9c)(9c – 5a) Question 21
Factorise:
a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3Solution 21
a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3
= [a (b – c)]3 + [b (c – a)]3 + [c (a – b)]3
Now, since, a (b – c) + b (c -a) + c (a – b)
= ab – ac + bc – ba + ca – bc = 0
So, a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3
= 3a (b – c) b (c – a) c (a – b)
= 3abc (a – b) (b – c) (c – a).Question 22(i)
Evaluate
(-12)3 + 73 + 53Solution 22(i)
Given equation is (-12)3 + 73 + 53
Now,
-12 + 7 + 5 = -12 + 12 = 0
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz
Hence,
(-12)3 + 73 + 53
= 3(-12)(7)(5)
= -1260 Question 22(ii)
Evaluate
(28)3 + (-15)3 + (-13)3Solution 22(ii)
Given equation is (28)3 + (-15)3 + (-13)3
Now,
28 + (-15) + (-13) = 28 – 28 = 0
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz
Hence,
(28)3 + (-15)3 + (-13)3
= 3(28)(-15)(-13)
= 16380 Question 23
Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a)Solution 23
L.H.S. = (a + b + c)3 – a3 – b3 – c3
= [(a + b) + c]3 – a3 – b3 – c3
= (a + b)3 + c3 + 3(a + b)c × [(a + b) + c] – a3 – b3 – c3
= a3 + b3 + 3ab(a + b) + c3 + 3(a + b)c × [(a + b) + c] – a3 – b3 – c3
= 3ab(a + b) + 3(a + b)c × [(a + b) + c]
= 3(a + b)[ab + c(a + b) + c2]
= 3(a + b)[ab + ac + bc + c2]
= 3(a + b)[a(b + c) + c(b + c)]
= 3(a + b)[(b + c)(a + c)]
= 3(a + b)(b + c)(c + a)
= R.H.S.Question 24
If a, b, c are all nonzero and a + b + c = 0, prove that
.Solution 24
Question 25
If a + b + c = 9 and a2 + b2 + c2 = 35, find the value of (a3 + b3 + c3 – 3abc).Solution 25
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
= (a + b + c)[(a2 + b2 + c2) – (ab + bc + ca)] ….(i)
Now,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ (9)2 = 35 + 2(ab + bc + ca)
⇒ 81 = 35 + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = 46
⇒ ab + bc + ca = 23
Substituting in (i), we get
a3 + b3 + c3 – 3abc = (9)[35 – 23] = 9 × 12 = 108
Exercise Ex. 3B
Question 1
Factorise:
9x2 – 16y2Solution 1
9x2 – 16y2
= (3x)2 – (4y)2
= (3x + 4y)(3x – 4y) Question 2
Factorise:
Solution 2
Question 3
Factorise:
81 – 16x2Solution 3
81 – 16x2
= (9)2 – (4x)2
= (9 – 4x)(9 + 4x) Question 4
Factorise:
5 – 20x2Solution 4
5 – 20x2
= 5(1 – 4x2)
= 5[(1)2 – (2x)2]
= 5[(1 – 2x)(1 + 2x)]
= 5(1 – 2x)(1 + 2x) Question 5
Factorise:
2x4 – 32Solution 5
2x4 – 32
= 2(x4 – 16)
= 2[(x2)2 – (4)2]
= 2[(x2 – 4)(x2 + 4)]
= 2[(x2 – 22)(x2 + 4)]
= 2[(x – 2)(x + 2)(x2 + 4)]
= 2(x – 2)(x + 2)(x2 + 4) Question 6
Factorise:
3a3b – 243ab3Solution 6
3a3b – 243ab3
= 3ab (a2 – 81 b2)
= 3ab [(a)2 – (9b)2]
= 3ab (a + 9b) (a – 9b)Question 7
Factorise:
3x3 – 48xSolution 7
3x3 – 48x
= 3x (x2 – 16)
= 3x [(x)2 – (4)2]
= 3x (x + 4) (x – 4)Question 8
Factorise:
27a2 – 48b2Solution 8
27a2 – 48b2
= 3 (9a2 – 16b2)
= 3 [(3a)2 – (4b)2]
= 3(3a + 4b) (3a – 4b)Question 9
Factorise:
x – 64x3Solution 9
x – 64x3
= x (1 – 64x2)
= x[(1)2 – (8x)2]
= x (1 + 8x) (1 – 8x)Question 10
Factorise:
8ab2 – 18a3Solution 10
8ab2 – 18a3
= 2a (4b2 – 9a2)
= 2a [(2b)2 – (3a)2]
= 2a (2b + 3a) (2b – 3a)Question 11
Factorise:
150 – 6x2Solution 11
150 – 6x2
= 6 (25 – x2)
= 6 (52 – x2)
= 6 (5 + x) (5 – x)Question 12
Factorise:
2 – 50x2Solution 12
2 – 50x2
= 2 (1 – 25x2)
= 2 [(1)2 – (5x)2]
= 2 (1 + 5x) (1 – 5x)Question 13
Factorise:
20x2 – 45Solution 13
20x2 – 45
= 5(4x2 – 9)
= 5 [(2x)2 – (3)2]
= 5 (2x + 3) (2x – 3)Question 14
Factorise:
(3a + 5b)2 – 4c2Solution 14
(3a + 5b)2 – 4c2
= (3a + 5b)2 – (2c)2
= (3a + 5b – 2c)(3a + 5b + 2c) Question 15
Factorise:
a2 – b2 – a – bSolution 15
a2 – b2 – a – b
= a2 – b2 – (a + b)
= (a – b)(a + b) – (a + b)
= (a + b)(a – b – 1) Question 16
Factorise:
4a2 – 9b2 – 2a – 3bSolution 16
4a2 – 9b2 – 2a – 3b
= (2a)2 – (3b)2 – (2a + 3b)
= (2a – 3b)(2a + 3b) – (2a + 3b)
= (2a + 3b)(2a – 3b – 1) Question 17
Factorise:
a2 – b2 + 2bc – c2Solution 17
a2 – b2 + 2bc – c2
= a2 – (b2 – 2bc + c2)
= a2 – (b – c)2
= [a – (b – c)][a + (b – c)]
= (a – b + c)(a + b – c) Question 18
Factorise:
4a2 – 4b2 + 4a + 1Solution 18
4a2 – 4b2 + 4a + 1
= (4a2 + 4a + 1) – 4b2
= [(2a)2 + 2 × 2a × 1 + (1)2] – (2b)2
= (2a + 1)2 – (2b)2
= (2a + 1 – 2b)(2a + 1 + 2b)
= (2a – 2b + 1)(2a + 2b + 1) Question 19
Factorise:
a2 + 2ab + b2 – 9c2Solution 19
a2 + 2ab + b2 – 9c2
= (a + b)2 – (3c)2
= (a + b + 3c) (a + b – 3c)Question 20
Factorise:
108a2 – 3(b – c)2Solution 20
108a2 – 3(b – c)2
= 3 [(36a2 – (b -c)2]
= 3 [(6a)2 – (b – c)2]
= 3 (6a + b – c) (6a – b + c)Question 21
Factorise:
(a + b)3 – a – bSolution 21
(a + b)3 – a – b
= (a + b)3 – (a + b)
= (a + b) [(a + b)2 – 12]
= (a + b) (a + b + 1) (a + b – 1)Question 22
Factorise:
x2 + y2 – z2 – 2xySolution 22
x2 + y2 – z2 – 2xy
= (x2 + y2 – 2xy) – z2
= (x – y)2 – z2
= (x – y – z)(x – y + z) Question 23
Factorise:
x2 + 2xy + y2 – a2 + 2ab – b2Solution 23
x2 + 2xy + y2 – a2 + 2ab – b2
= (x2 + 2xy + y2) – (a2 – 2ab + b2)
= (x + y)2 – (a – b)2
= [(x + y) – (a – b)][(x + y) + (a – b)]
= (x + y – a + b)(x + y + a – b) Question 24
Factorise:
25x2 – 10x + 1 – 36y2Solution 24
25x2 – 10x + 1 – 36y2
= (25x2 – 10x + 1) – 36y2
= [(5x)2 – 2(5x)(1) + (1)2] – (6y)2
= (5x – 1)2 – (6y)2
= (5x – 1 – 6y)(5x – 1 + 6y) Question 25
Factorise:
a – b – a2 + b2Solution 25
a – b – a2 + b2
= (a – b) – (a2 – b2)
= (a – b) – (a – b) (a + b)
= (a – b) (1 – a – b)Question 26
Factorise:
a2 – 4ac + 4c2 – b2Solution 26
a2 – 4ac + 4c2 – b2
= a2 – 4ac + 4c2 – b2
= a2 – 2 a 2c + (2c)2 – b2
= (a – 2c)2 – b2
= (a – 2c + b) (a – 2c – b)Question 27
Factorise:
9 – a2 + 2ab – b2Solution 27
9 – a2 + 2ab – b2
= 9 – (a2 – 2ab + b2)
= 32 – (a – b)2
= (3 + a – b) (3 – a + b)Question 28
Factorise:
x3 – 5x2 – x + 5Solution 28
x3 – 5x2 – x + 5
= x2 (x – 5) – 1 (x – 5)
= (x – 5) (x2 – 1)
= (x – 5) (x + 1) (x – 1)Question 29
Factorise:
1 + 2ab – (a2 + b2)Solution 29
1 + 2ab – (a2 + b2)
= 1 – (a2 + b2 – 2ab)
= (1)2 – (a – b)2
= [1 – (a – b)][1 + (a – b)]
= (1 – a + b)(1 + a – b) Question 30
Factorise:
9a2 + 6a + 1 – 36b2Solution 30
9a2 + 6a + 1 – 36b2
= (9a2 + 6a + 1) – 36b2
= [(3a)2 + 2(3a)(1) + (1)2] – (6b)2
= (3a + 1)2 – (6b)2
= (3a + 1 – 6b)(3a + 1 + 6b) Question 31
Factorise:
x2 – y2 + 6y – 9Solution 31
x2 – y2 + 6y – 9
= x2 – (y2 – 6y + 9)
= x2 – (y2 – 2 y 3 + 32)
= x2 – (y – 3)2
= [x + (y – 3)] [x – (y – 3)]
= (x + y – 3) (x – y + 3)Question 32
Factorise:
4x2 – 9y2 – 2x – 3ySolution 32
4x2 – 9y2 – 2x – 3y
= (2x)2 – (3y)2 – (2x + 3y)
= (2x + 3y) (2x – 3y) – (2x + 3y)
= (2x + 3y) (2x – 3y – 1)Question 33
Factorise:
9a2 + 3a – 8b – 64b2Solution 33
9a2 + 3a – 8b – 64b2
= 9a2 – 64b2 + 3a – 8b
= (3a)2 – (8b)2 + (3a – 8b)
= (3a + 8b) (3a – 8b) + (3a – 8b)
= (3a – 8b) (3a + 8b + 1)Question 34
Factorise:
Solution 34
Question 35
Factorise:
Solution 35
Question 36
Factorise:
Solution 36
Question 37
Factorise:
x8 – 1Solution 37
x8 – 1
= (x4)2 – (1)2
= (x4 – 1)(x4 + 1)
= [(x2)2 – (1)2)(x4 + 1)
= (x2 – 1)(x2 + 1)(x4 + 1)
= (x – 1)(x + 1)(x2 + 1)[(x2)2 + (1)2 + 2x2 – 2x2
= (x – 1)(x + 1)(x2 + 1)[(x2)2 + (1)2 + 2x2) – 2x2
Question 38
Factorise:
16x4 – 1Solution 38
16x4 – 1
= (4x2)2 – (1)2
= (4x2 – 1)(4x2 + 1)
= [(2x)2 – (1)2](4x2 + 1)
= (2x – 1)(2x + 1)(4x2 + 1) Question 39
Factorise:
81x4 – y4Solution 39
81x4 – y4
= (9x2)2 – (y2)2
= (9x2 – y2)(9x2 + y2)
= [(3x)2 – y2](9x2 + y2)
= (3x – y)(3x + y)(9x2 + y2) Question 40
Factorise:
x4 – 625Solution 40
x4 – 625
= (x2)2 – (25)2
= (x2 + 25) (x2 – 25)
= (x2 + 25) (x2 – 52)
= (x2 + 25) (x + 5) (x – 5)
Exercise Ex. 3A
Question 1
Factorise:
9x2 + 12xySolution 1
9x2 + 12xy = 3x (3x + 4y)Question 2
Factorise:
18x2y – 24xyzSolution 2
18x2y – 24xyz = 6xy (3x – 4z)Question 3
Factorise:
27a3b3 – 45a4b2Solution 3
27a3b3 – 45a4b2 = 9a3b2 (3b – 5a)Question 4
Factorise:
2a (x+ y) – 3b (x + y)Solution 4
2a (x + y) – 3b (x + y) = (x + y) (2a – 3b)Question 5
Factorise:
2x (p2 + q2) + 4y (p2 + q2)Solution 5
2x (p2 + q2) + 4y (p2 + q2)
= (2x + 4y) (p2 + q2)
= 2(x+ 2y) (p2 + q2)Question 6
Factorise:
x (a – 5) + y (5 – a)Solution 6
x (a – 5) + y (5 – a)
= x (a – 5) + y (-1) (a – 5)
= (x – y) (a – 5)Question 7
Factorise:
4 (a + b) – 6 (a + b)2Solution 7
4 (a + b) – 6 (a + b)2
= (a + b) [4 – 6 (a + b)]
= 2 (a + b) (2 – 3a – 3b)
= 2 (a + b) (2 – 3a – 3b)Question 8
Factorise:
8 (3a – 2b)2 – 10 (3a – 2b)Solution 8
8 (3a – 2b)2 – 10 (3a – 2b)
= (3a – 2b) [8(3a – 2b) – 10]
= (3a – 2b) 2[4 (3a – 2b) – 5]
= 2 (3a – 2b) (12 a – 8b – 5)Question 9
Factorise:
x (x + y)3 – 3x2y (x + y)Solution 9
x (x + y)3 – 3x2y (x + y)
= x (x + y) [(x + y)2 – 3xy]
= x (x + y) (x2 + y2 + 2xy – 3xy)
= x (x + y) (x2 + y2 – xy)Question 10
Factorise:
x3+ 2x2 + 5x + 10Solution 10
x3+ 2x2 + 5x + 10
= x2 (x + 2) + 5 (x + 2)
= (x2 + 5) (x + 2)Question 11
Factorise:
x2 + xy – 2xz – 2yzSolution 11
x2 + xy – 2xz – 2yz
= x (x + y) – 2z (x + y)
= (x+ y) (x – 2z)Question 12
Factorise:
a3b – a2b + 5ab – 5bSolution 12
a3b – a2b + 5ab – 5b
= a2b (a – 1) + 5b (a – 1)
= (a – 1) (a2b + 5b)
= (a – 1) b (a2 + 5)
= b (a – 1) (a2 + 5)Question 13
Factorise:
8 – 4a – 2a3 + a4Solution 13
8 – 4a – 2a3 + a4
= 4(2 – a) – a3 (2 – a)
= (2 – a) (4 – a3)Question 14
Factorise:
x3 – 2x2y + 3xy2 – 6y3Solution 14
x3 – 2x2y + 3xy2 – 6y3
= x2 (x – 2y) + 3y2 (x – 2y)
= (x – 2y) (x2 + 3y2)Question 15
Factorise:
px – 5q + pq – 5xSolution 15
px + pq – 5q – 5x
= p(x + q) – 5 (q + x)
= (x + q) (p – 5)Question 16
Factorise:
x2 + y – xy – xSolution 16
x2 – xy + y – x
= x (x – y) – 1 (x – y)
= (x – y) (x – 1)Question 17
Factorise:
(3a – 1)2 – 6a + 2Solution 17
(3a – 1)2 – 6a + 2
= (3a – 1)2 – 2 (3a – 1)
= (3a – 1) [(3a – 1) – 2]
= (3a – 1) (3a – 3)
= 3(3a – 1) (a – 1)Question 18
Factorise:
(2x – 3)2 – 8x + 12Solution 18
(2x – 3)2 – 8x + 12
= (2x – 3)2 – 4 (2x – 3)
= (2x – 3) (2x – 3 – 4)
= (2x – 3) (2x – 7)Question 19
Factorise:
a3 + a – 3a2 – 3Solution 19
a3 + a – 3a2 – 3
= a(a2 + 1) – 3 (a2 + 1)
= (a – 3) (a2 + 1)Question 20
Factorise:
3ax – 6ay – 8by + 4bxSolution 20
3ax – 6ay – 8by + 4bx
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b)Question 21
Factorise:
abx2 + a2x + b2x +abSolution 21
abx2 + a2x + b2x +ab
= ax (bx + a) + b (bx + a)
= (bx + a) (ax + b)Question 22
Factorise:
x3 – x2 + ax + x – a – 1Solution 22
x3 – x2 + ax + x – a – 1
= x3 – x2 + ax – a + x – 1
= x2 (x – 1) + a (x – 1) + 1 (x – 1)
= (x – 1) (x2 + a + 1)Question 23
Factorise:
2x + 4y – 8xy – 1Solution 23
2x + 4y – 8xy – 1
= 2x – 1 – 8xy + 4y
= (2x – 1) – 4y (2x – 1)
= (2x – 1) (1 – 4y)Question 24
Factorise:
ab (x2 + y2) – xy (a2 + b2)Solution 24
ab (x2 + y2) – xy (a2 + b2)
= abx2 + aby2 – a2xy – b2xy
= abx2 – a2xy + aby2 – b2xy
= ax (bx – ay) + by(ay – bx)
= (bx – ay) (ax – by)Question 25
Factorise:
a2 + ab (b + 1) + b3Solution 25
a2 + ab (b + 1) + b3
= a2 + ab2 + ab + b3
= a2 + ab + ab2 + b3
= a (a + b) + b2 (a + b)
= (a + b) (a + b2)Question 26
Factorise:
a3 + ab (1 – 2a) – 2b2Solution 26
a3 + ab (1 – 2a) – 2b2
= a3 + ab – 2a2b – 2b2
= a (a2 + b) – 2b (a2 + b)
= (a2 + b) (a – 2b)Question 27
Factorise:
2a2 + bc – 2ab – acSolution 27
2a2 + bc – 2ab – ac
= 2a2 – 2ab – ac + bc
= 2a (a – b) – c (a – b)
= (a – b) (2a – c)Question 28
Factorise:
(ax + by)2 + (bx – ay)2Solution 28
(ax + by)2 + (bx – ay)2
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2abxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + b2x2 + b2y2 + a2y2
= x2 (a2 + b2) + y2(a2 + b2)
= (a2 + b2) (x2 + y2)Question 29
Factorise:
a (a + b – c) – bcSolution 29
a (a + b – c) – bc
= a2 + ab – ac – bc
= a(a + b) – c (a + b)
= (a – c) (a + b)Question 30
Factorise:
a(a – 2b – c) + 2bcSolution 30
a(a – 2b – c) + 2bc
= a2 – 2ab – ac + 2bc
= a (a – 2b) – c (a – 2b)
= (a – 2b) (a – c)Question 31
Factorise:
a2x2 + (ax2 + 1)x + aSolution 31
a2x2 + (ax2 + 1)x + a
= a2x2 + ax3 + x + a
= ax2 (a + x) + 1 (x + a)
= (ax2 + 1) (a + x)Question 32
Factorise:
ab (x2 + 1) + x (a2 + b2)Solution 32
ab (x2 + 1) + x (a2 + b2)
= abx2 + ab + a2x + b2x
= abx2 + a2x + ab + b2x
= ax (bx + a) + b (bx + a)
= (bx + a) (ax + b)Question 33
Factorise:
x2 – (a + b) x+ abSolution 33
x2 – (a + b) x+ ab
= x2 – ax – bx + ab
= x (x – a) – b(x – a)
= (x – a) (x – b)Question 34
Factorise:
Solution 34
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