Gujarat Board Textbook Solutions Class 9 Maths Chapter 2 Polynomials

 Gujarat Board Textbook Solutions Class 9 Maths Chapter 2 Polynomials

GSEB Solutions Class 9 Maths Chapter 2 Polynomials

Ex 2.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x2 – 3x + 7
(ii) y2 + √2
(iii) 3√t + t√2
(iv) y + 1/2
(v) x10 + y3 + t50
Solution:
(i) 4x2 – 3x + 7
This is a polynomial in one variable x whose powers are non-negative integers. So this polynomial is in one variable.

(ii) y2 + y√2
This polynomial is in one variable y whose exponents are in whole numbers. So this polynomial is in one variable.

(iii) 3√t + t√2
This is not a polynomial because exponent of t is 1/2 which is not a whole number.

(iv) y + 2/y This is not a polynomial because exponent 2/y of i.e., 2y-1 is negative which is not a whole number. So this is not a polynomial in one variable.

(v) x10 + y3 + t50
This polynomial is in three variables x, y and t. Their exponents are in whole numbers. So this is a polynomial in three variables.

Question 2.
Write the coefficients of x2 in each of the following:

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
5x35 + 9 and 7x100

Question 4.
Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x
(ii) 4 – y2
(iii) 5t – √7
(iv) 3
Solution:
(i) 5x3 + 4x2 + 7x
The term 5x3 is highest power in x whose power is 3. Therefore degree of polynomial is 3.

(ii) 4 – y2
-y2 is highest power iny which has power 2.
S0 degree of polynomial is 2.

(iii) 5t – √7
5t has highest power in t which has power 1. So degree of polynomial is 1.

(iv) 3 has no variable i.e., exponent of variable is 0. Therefore degree of constant polynomial is 0.

Question 5.
Classify the following as linear, quadratic and cubic polynomials.
(i) x2 + x
(ii) x – x3
(iii) y + y2 + 4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7x3
Solution:
(i) x2 + x Quadratic (Highest degree 2)
(ii) x – x3 Cubic (Highest degree 3)
(iii) y + y2 + 4 Quadratic (Highest degree 2)
(iv) 1 + x Linear (Highest degree 1)
(v) 3t Linear (Highest degree 1)
(vi) r2 Quadratic (Highest degree 2)
(vii) 7x3 Cubic (Highest degree 3)

Ex 2.2

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
Let p(x) = 5x – 4x2 + 3
(i) Substituting x = 0 in p(x)
p(0) = 5 x (0) – 4 x (0)2 + 3
p(0) = 3

(ii) Substituting x = – 1 in p(x)
p(- 1) = 5 x (- 1) – 4(- 1)2 + 3
= – 5 – 4 + 3 = – 9 + 3 = – 6

(iii) Substituting x = 2 in p(x)
p(2) = 5 x 2 – 4(2)2 + 3
= 10 – 4 x 4 + 3 = 13 – 16 = – 3

Question 2.
Find p(0), p(1) and p( 2) for each of the following polynomials:
(i) p(y) = y2 – y + 1
(ii) p(t) = 2 + t + 2t2 – t3
(iii) p(x) = x3
(iv) p(x) = (x – 1) (x + 1)
Solution:
(i) p(y) = y2 – y + 1
∴ p(0) = (0)2 – (0) + 1 = 1
p(y) = y2 – y + 1
p(1) = (1)2 – (1) + 1 = 1 – 1 = 1
p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 2 + 1 = 3

(ii) p(t) = 2 + t + 2t2 – t3
∴ p(0) = 2 + (0) + 2(0)2 – (0)3 = 2
p(1) = 2 + (1) + 2(1)2 – (1)3
= 2 + 1 + 2 – 1 = 4
p(2) = 2 + 2 + 2(2)2 – 23
= 4 + 2 x 4 – 8
= 4 + 8 – 8 = 4

(iii) p(x) = x3
∴ p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8

(iv) p(x) = (x – 1) (x + 1)
∴ p(0) = (0 – 1) (0 + 1) = (- 1) x (1) = – 1
p(1) = (1 – 1) (1 + 1) = 0 x 2 = 0
p(2) = (2 – 1) (2 + 1) = 1 x 3 = 3

Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them.

(iii) p(x) = x2 – 1, x = 1, – 1
p(1) = (1)2 – 1, 1 = 1 – 1 = 0
p(- 1) = (- 1)2 – 1 = 1 – 1 = 0
∴ x = 1, and x = – 1 are zeroes

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
p(- 1) = (- 1 + 1) (- 1 – 2) .
= 0 x (- 3) = 0
p(2) = (2 + 1) (2 – 2) = 3 x (0)
∴ p(2) = 0
Hence x = – 1, and x = 2 are zeroes of p(x).

(v) p(x) = x2, x = 0
p(0) = (0)2 = 0
∴ x = 0 is a zero of p(x)

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Solution:
(i) p(x) = x + 5
Let p(x) = 0
⇒ x + 5 = 0
⇒ x = – 5
∴ x = – 5 is zero of polynomial p(x)

(ii) p(x) = x – 5
Let p(x) = 0
⇒ x – 5 = 0
⇒ x = 5
∴ x = 5 is zero of polynomial p(x)

(iii) p(x) = 2x + 5
Let p(x) = 0
⇒ 2x + 5 = 0
⇒ 2x = 5
∴ x = −5/2 is zero of polynomial p(x)

(iv) p(x) = 3x – 2
Let p(x) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
∴ x = 2/3 is zero of polynomial p(x)

(v) p(x) = 3x
Let p(x) = 0
⇒ 3x = 0
⇒ x = 0
∴ x = 0 is zero of polynomial p(x)

(vi) p(x) = ax, a ≠ 0
Let p(x) = 0
⇒ ax = 0
⇒ x = 0/a
⇒ x = 0
∴ x = 0 is zero of polynomial p(x)

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Let p(x) = 0
⇒ cx + d = 0
⇒ cx = – d
⇒ x = −d/c
∴ x = −d/c is zero of polynomial p(x)

Ex 2.3

Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
1. x + 1
2. x – 1/2
3. x
4. x + π
5. 5 + 2x
Solutiobn:
p(x) = x3 + 3x2 + 3x + 1
1. Let x + 1 = 0
x = – 1
Then p(-l) = (-1)3 + 3(-1)2 + 3(-1) + 1
= -1 + 3 – 3 + 1= 0
∴ Remainder = 0

3. Let x = 0
Then p(0) = 03 + 3(0)2 + 3(0) + 1 = 1
∴ Remainder = 1

4. Let x + π = 0
x = – π
Then p(- π) = (-π)3 + 3(-π)2 + 3π + 1
= -π3 + 3π2 + 3π + 1
∴ Remainder = -π3 + 3π2 + 3π + 1

Question 2.
Find the remainder when x3 – ax3 + 6x – a is divided by x – a.
Solution:
Let p(x) = x3 – ax2 + 6x – a
and x – a = 0
= x = a
∴ p(a) = a3 – a x a2 + 6a – a
= a3 – a3 + 6a – a
p(a) = 5a
Hence remainder = 5a

Question 3.
Check whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
7 + 3x will be a factor of polynomial 3x3 + 7x if we divide 3x3 + 7x by 7 + 3x and it leaves no remainder.
Let p(x) = 3x3 + 7x
and 7 + 3x = 0

Ex 2.4

Question 1.
Determine which of the following polynomials has x + 1 a factor:
1. x3 + x2 + x + 1
2. x4 + x2+ x2 + x + 1
3. x4 + 3x3 + 3x2 + x + 1
4. x3 – x2 – (2 + √2)x + √2
Solution:
1.  Let p(x) = x3 + x2 + x + 1
x + 1 = 0
x = -1 (If x + 1 is a factor of p(x) then x + 1 will be equal to zero)
p(-1) = (-1)3 + (-1)2 + (-1) + 1
= -1 + 1 – 1 + 1 = 0
∴ By factor theorem x + 1 is a factor of x3 + x2 + x + 1.

2. Let p(x) = x4 + x3 + x2 + x + 1
⇒ x + 1 = 0
x = -1
p(-1) = (-1)4 + (-1)3 + (-l)2 + (-l) + 1
= 1 – 1 + 1 – 1 + 1 = 3 – 2
⇒ p(-1) = 1
∴ 1 ≠ 0
Hence by factor theorem x + 1 is not a factor of x4 + x3 + x2 + x + 1.

3. Let p(x) = x4 + 3x3 + 3x2 + x + 1
x + 1 = 0
⇒ x = -1
p(-1) = (-1)4 + 3(-1)3 + 3(-1)2 + (-1) + 1
p(-1) = 1 – 3 + 3
p(-1) = 1
∴ 1≠ 0
Hence by factor theorem x + 1 is not a factor of x4 + 3x3 + x2 + x + 1.

Question 2.
Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
1. p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1
2. p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
3. p(x) = x3 – 4x2 + x + 6, g(x) = x – 3
Solution:
1. p(x) = 2 x3 + x2 – 2x – 1
g(x) = x + 1
g(x) = 0
⇒ x + 1 = 0
⇒ x = -1
p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1
= -2 + 1 + 2 – 1
⇒ p(-1) = 0
∴ By factor theorem g(x) is a factor of p(x).

2. p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
g(x) = 0
x + 2 = 0
x = -2
p(-2) = (-2)3 + 3(-2)2 + 3(-2) + 1
= p(-2) = – 8 + 3 x 4 – 6 + 1
= p(-2) = – 8 + 12 – 6 + 1
= p(-2) = -1
∴ -1 ≠ 0
Hence by factor theorem g(x) is not a factor of p(x).

3. p(x) = x3 – 4x2 + x + 6, g(x) = x – 3
g(x) = 0
⇒ x – 3 = 0
⇒ x = 3
p(3) = 33 – 4(3)2 + 3 + 6
⇒ p(3) = 27 – 4 x 9 + 9
⇒ p(3) = 27 – 36 + 9
⇒ p(3) = 36 – 36
p(3) = 0
Hence by factor theorem g(x) is a factor of p(x).

Question 3.
Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
1. p(x) = x2 + x + k
2. p(x) = 2x2 + kx + √2
3. p(x) = kx2 – √2 x + 1
4. p(x) = kx2 – 3x + k
Solution:
1. p(x) = x2 + x + k
x – 1 will be a factor p(x) therefore p(1) = 0.
By factor theorem
:. p(1) = 12 + 1 + k
0 = 2 + k = k = -2

4. p(x) = kx2 – 3x + k
x – 1 will be a factor of p(x),
therefore, p(1) = 0.
By factor theorem,
p(1) = k(1)2 – 3(1) + k
0 = k – 3 + k
0 = 2k – 3 = k = 3/2

Question 4.
Factorise: 2
1. 12 x2 – 7x + 1
2. 2x2 + 7x + 3
3. 6x2 + 5x – 6
4. 3x2 – x – 4
Solution:
1. 12 x2 – 7x + 1 (a x c = 12 x 1)
= 12 x2 – 4x – 3x + 1 (12 = 4 x 3)
= 4 x (3x -1) -1(3x – 1.)
= (3x – 1)(4x – 1)

2. 2x2 + 7x + 3 (a x c = 2 x 3)
-2x2 + 6x +x + 3 (6 = 6 x 1)
= 2x (x + 3) + 1(x + 3)
= (x + 3) (2x + 1)

3. 6x2 + 5x – 6 (a x c = 6 x 6 = 36)
= 6x2 + 9x – 4x – 6 (36 = 9 x 4)
= 3x(2x + 3) -2(2x + 3)
= (2x + 3)(3x – 2)

4. 3x2 – x – 4 (a x c = 3 x 4)
= 3x2 – 4x + 3x – 4 (12 = 4 x 3)
= x (3x – 4)+ 1(3x – 4)
= (3x – 4) (x + 1)

Question 5.
Factorise:
1. x3 – 2x2 – x + 2
2. x3 – 3x2 – 9x – 5
3.  x3+ 13x2 + 32x + 20
5. 2y2 + y2 – 2y – 1
Solution:
1. Let p(x) = x3 – 2x2 – x + 2
Here constant form is 2 and its factors can be ±1, ±2.
By trial method we find factor of p(x).
Let x – 1 be a factor of p(x).
∴ x – 1 = 0
⇒ x = 1
p(1) = 13 – 2(1)3 – 1 + 2
= 1 – 2 – 1 + 2 = 3 – 3
⇒ p(1) = 0
∴ By factor theorem x – 1 is a factor of p(x).
Now x3 – 2x3 – x + 2
= x2 – x2 + x – 2x + 2
= x(x – 1) -x(x – 1)-2(x – 1)
= (x – 1)(x2 – x – 2)
= (x – 1)[x2 – 2x + x – 2]
= (x – 1)[x(x – 2) + 1(x – 2)]
= (x – 1) (x – 2) (x + 1)

2. Let p(x) = x3 – 3x2 – 9x – 5
Here constant term is 5 and its factors can be ±1, ±5.
By trial method, we find factor of p(x).
Let x – 1 be a factor of p(x).
∴ x – 1= 0
⇒ x = 1
p(1) = 13 – 3(1)2 – 9(1) – 5
= 1 – 3 – 9 – 5 = -16
But -16≠0
∴ x – 1 is not a factor of p(x).
Now let x + 1 be a factor of p(x).
∴ x + 1 = 0
⇒ x = -1
p(-1) = (-1)3 – 3(-1)2 – 9(-1) – 5
= -1 – 3 + 9 – 5 = -9 + 9
p(-1) = 0
∴ x + 1 is a factor of p(x).

Now x3 – 3x2 – 9x – 5
= x3 + x3 – 4x3 – 4x – 5x – 5
= x2(x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x2 – 4x – 5)
= (x + 1)(x2 – 5x + x – 51
= (x + 1) [x(x – 5)+ 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
= (x + 1)(x + 1)(x – 5)

3. Let p(x) = x3 + 13x2 + 32x + 20 Here constant term is 20 and its factors can be ±1, ±2, ±4, ±5, ±10, ±20
By trial method, we find the factor of p(x).
Let x + 1 be a factor of p(x).
∴ x + 1 = 0
x = – 1
p(-1) = (-1) + 13(-1)2 + 32(-1) + 20
= -1 + 13 – 32 + 20 = 33 – 33
p(-1) = 0
x + 1 is a factor of p(x).
Hence X3 + 13x2+ 32x + 20
= x3 +x2 + 12x2 + 12x + 20x + 20
= x2(x + 1) + 12x(x + 1) + 20(x + 1)
= (x + 1)(x2 + 12x + 20)
= (x + 1)[x2 + 10 x + 2x + 20]
= (x + 1)[x(x + 10)+ 2(x + 10)]
= (x + 1)(x+1o)(x + 2)
= (x + 1)(x + 2)(x + 10)

4. Let p(y) = 2y2 + y2 – 2y – 1
By trial method, we find the factor of p(y).
Let y – 1 be a factor of p(y).
y – 1 = 0 ⇒ y = 1
p(1) = 2(1)3 + 12 – 2 x 1 – 1
= 2 + 1 – 2 – 1 = 3 – 3
p(1) = 0
∴ y – 1 is a factor of p(y).
Hence 2y3 + y2 – 2y – 1
2y3 – 2y2 + 3y2 – 3y + y – 1
= 2y2(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)[2y2 + 3y + 11]
= (y – 1) [2y2 + 2y + y + 1]
= (y -1)[2y(y + 1) +1 (y + 1)]
= (y – 1)(y + 1)(2y + 1)

Ex 2.5

Question 1.
Use suitable identities to find the following products:
(i) (x + 4)(x + 10)
(ii) (x + 8)(x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y2 + 3/2 )(y2 – 3/2)
(v) (3 – 2x)(3 + 2x)
Solution:
(i) Using identity
(x + a)(x + b) = x2 + (a + b) x + ab, we get
= (x + 4)(x + 10) = x2 + (4 + 10) x + 4 x 10
= x2 + 14x + 40

(ii) Using identity
(x + a)(x + b) = x2 + (a + b) x + ab,we get
(x + 8)(x – 10)
= (x + 8) {x + (-10)}
= x2 + {8 + (-10)} x + 8 x(-10)
= x2 + {8 – 10} x – 80
= x2 – 2x – 80

(iii) Using identity
(x + a) (x + b) = x2 + (a + b)x + ab, we get
(3x + 4) (3x – 5)
= (3x)2+{4 + (-5)(3x) + 4x(-5)
= 9x2 + (-1)3x – 20
= 9x2 – 3x – 20

(iv) Using identity
(a + b)(a – b) = a2 – b2,we get
(y2 + 3/2) (y2 – 3/2) = (y2)2 – (3/2)2 = y – 9/4

(v) Using identity
(a – b)(a + b) = a2 – b2,weget
(3 – 2x)(3 + 2x) = 32 – (2x)2
= 9 – 4x2

Question 2.
Evaluate the following products without multiplying directly:
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
Solution:
(i) 103 x 107 = (100 + 3)(100 + 7)
Using identity
(x + a)(x + b) = x2 + (a + b)x + ab, we get (100 + 3) (100 + 7)
= 1002 + (3 + 7) x 100 + 3 x 7
= 10000 + 10 x 100 + 21
= 10000 + 1000 + 21
= 11021

(ii) 95 x 96 = (100 – 5)(100 – 4)
Using identity
(x + a)(x + b) = x2 + (a + b)x + ab,we get {100 + (-5)} {100 + (-4)}
= 1002 + {5 + (-4)} x 100 + (-5) (-4)
= 10000 + (-9) x 100 + 20
= 10000 – 900 + 20 = 9120

(iii) 104 x 96 = (100 + 4)(100 – 4)
Using identity
(a + b)(a – b) = a2 – b2, we get
(100 + 4)(100 – 4) = (100)2 – 42
= 10000 – 16 = 9984

Question 3.
Factorise the following using appropriate identities:

Question 4.
Expand each of the following using suitable identities:
(i) (x + 2y +4z)2
(ii) (2x – y + z)2
(iii) (-2x + 3y + 2z)2
(iv) (3a – 7b – c)2
(v)(-2x + 5y – 3z)2
(vi) [1/4 a – 1/2 b + 1]2
Solution:
Using identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2 x x x 2y + 2 + 2y + 4z + 2 + 4z + x
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

(ii) {2x + (-y) + z}2
= (2x)2 + (-y)2 + z2 + 2 x 2x x (-y) + 2(-y)z + 2 x z x 2x
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx

(iii) (- 2x + 3y + 2z)2
= (-2x)2 + (3y)2 + (2z)2 + (-2x) x 3y+ 2 x (3y) x (2z) + 2 x 2z x (-2x)
= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx

(iv) (3a – 7b – c)2 = {3a + (-7b) + (-c)}2
= (3a)2 + (-7b)2 + (-c)2 + 2 x 3a x (-7b) + 2 x (-7b)(-c) + 2x(-c) x 3a
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca

(v) (-2x + 5y – 3z)2
= {(-2x) + 5y + (-3z))2
= {(-2x)2 + (5y)2 + (-3z)2}
= (-2x)2 x (5y)2 + (-3z)2 + 2 x (-2x) x 5y + 2 x 5y x (-3z) + 2(-3z) x (-2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx

Question 5.
Factorise:
(i) 4x2 + 9y2 + 16z2 + l2xy – 24yz – 16xz
(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8zx
Solution:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2 x 2x x 3y + 2 x 3y x (-4z) + 2(- 4z) x (2x)
[∴ a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2]
= [2x + 3y + (-4z)]2 = (2x + 3y – 4z)2
= (2x + 3y – 4z) (2x + 3y – 4z)

Question 6.
Write the following cubes in expanded form:
(i) (2x + 1)3
(ii) (2a – 3b)3
(iii) (3/2 x + 1)3
(iv) (x – 2/3 y)3
Solution:
(i) Using identity
(a + b)3 = a3 + b3 + 3ab (a + b), we get
(2x + 1)3 = (2x)3 + 13 + 3 x 2x x 1(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 1 + 12x3 + 6x
= 8x3 + 12x3 + 6x + 1

(ii) Using identity
(a – b)3 = a3 – b3 – 3ab (a – b), we get
(2a – 3b)3 = (2a)3 – (3b)3 – 3 x 2a x 3b(2a – 3b)
= 8a3 – 27b3 – l8ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2

Question 7.
Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solution:
(i) (99)3 = (100 – 1)3
Using identity
(a – b)3 = a3 – b3 – 3ab(a – b), we get
(100 – 1) = 100 – 1 – 3 x 100 x 1(100 – 1)
= 1000000 – 1 – 300(100 – 1)
= 1000000 – 1 – 30000 + 300
= 970299

(ii) (102)3 = (100 + 2)
Using identity
(a + b)3 = a3 + b3 + 3ab(a + b),we get
(100+2)3 = 1003 + 23 + 3 x 100 x 2(100+2)
= 1000000 + 8 + 600 (100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208

(iii) (998)3 = (1000 – 2)3
Using identity
(a – b)3 = a3 – b3 – 3ab(a – b), we get (1000 – 2)3
= 1oo03 – 23 – 3 x 1000 x 2(1000 – 2)
1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 + 12000
= 994011992

Question 8.
Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3 – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b + 108ab2
(v) 27p3 – 1/216 – 9/2 p2 + 1/4p
Solution:
(i) 8a3 + b3 + 12a2b + 6ab2
= (2a)3 + b3 + 3 x 2a x b(2a + b)
= (2a + b)3
[∴ a3 + b3 + 3ab(a + b) = (a + b)3]
= (2a + b)(2a + b)(2a + b)

(ii) 8a3 – b3 – 12a2b +6ab2
= (2a)3 – b3 – 3 x 2a x b (2a – b)
= (2a – b)3
[∴ a3 – b3 – 3ab(a – b) = (a – b)3]
= (2a – b)(2a – b)(2a – b)

(iii) 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3 x 3 x 5a (3 – 5a)
= (3 – 5a)3
[∴ a3 – b3 – 3ab(a – b) = (a – b)3]
= (3 – 5a)(3 – 5a)(3 – 5a)

(iv) 64a3 – 27b3 – 14a2b + 108ab2
= (4a)3 – (3b)3 – 3 x 4a x 3b(4a – 3b)
=(4a – 3b)3
[∴ a3 – b3 – 3ab(a – b) = (a – b)3]
= (4a – 3b)(4a – 3b)(4a – 3b)

Question 9.
Verify:
(i) x3 + y3 = (x + y)(x2 – xy + y2)
(ii) x3 – y3 = (x – y)(x2 + xy + y2)
Solution:
(i) We know that
(x + y)3 = x3 + y3 + 3xy (x + y)
x3 + y3 (x + y)3 + 3xy(x+y)
= x3 + y3 = (x + y){(x + y)2 – 3xy)
= x3 + y3 = (x + y){x2 + y2 + 2xy – 3xy)
x3 + y3 = (x + y){x2 + y2 — xy)
=> x3 + y3 = (x + y)(x2 – xy + y2)

(ii) We know that
(x – y)3 =x3 – y3 – 3xy(x – y)
x3 – y3 = (x – y)3 + 3xy(x – y)
x3 – y = (x – y){(x – y)2 + 3xy}
⇒ x3 – y3 = (x – y)(x2 + y2 – 2xy + 3xy)
x3 – y3 = (x – y)(x2 + xy + y2)

Question 10.
Factorise each of the following:
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
Solution:
(i) 27 y3 + 125 z3 = (3y)3 + (5z)3
Using identity
a3 + b3 (a + b) (a2 – ab + b2), we get (3y)3 + (5z)3
= (3y + 5z) {(3y)2 – 3y x 5z + (5z)2)
= 27 y3 + 125 z3
= (3y + 5z)(9y2 – 15yz + 25z2)

(ii) 64m3 – 343n3 = (4m)3 – (7n)3
Using identity
a3 – b3 = (a – b) (a2 + ab + b2), we get
(4m)3 – (7n)3
= (4m – 7n){(4m)2 + 4m x 7n + (7n)2}
64m3 – 343n3
= (4m – 7n) (16m2 + 28mn + 49n2)

Question 11.
Factorise: 27x3 + y3 + z3 – 9xyz
Solution:
27 x3 + y3 + z 3 – 9xyz
= (3x)3 + y3 + z3 – 3 x 3x x y x z
Using identity
a3 + b3 + c3 – 3abc
= (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
27 x3 + y3 + z3 – 9xyz
= (3x +y – z){(3x)2 + y2 + z2 – 3x x y – yz – z x 3x)
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx)

Question 12.
Verify that x3 + y3 + z3 – 3xyz = (x + y + z)((x – y)2 + (y – z)2+(z – x)3]
Solution:
Using identity
a3 + b3 + c3 – 3abc
= (a + b + c) (a2 + b2 + c2 -ab – bc – cu),we get x3 + y3 + z3 – 3xyz
= (x + y + z)[x2 + y2 + z2 – xy – yz – zx]
= 1/2 (x + y + z)[2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]
= 1/2 (x + y + z)[x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx]
1/2 (x + y + z) [(x – y)2 + (y – z)2 + z – x)2]

Question 13.
If x + y + z = 0, show that x2 + y2 + z2 = 3xyz.
Solution:
We know that
x2 + y2 + z2 – 3xyz
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
But we have x + y + z = 0
∴ x2 + y2 + z2 – 3xyz
= 0 x (x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = 0
∴ x3 + y3 + z3 = 3xyz

Question 14.
Without actually calculating the cubes, find the value of each of the following:
(i) (-12)3 + (7)3 + (5)3
(ii) (28)3 + (-15)3 + (-13)3
Solution:
(i) Let a = -12, b = 7 and c = 5
a + b + c = -12 + 7 + 5 = – 12 + 12
We know that ifa + b + c = 0, then
a3 + b3 + c3 = 3abc
=(-12) + (7)3 + (5)3 = 3 x(-12) x 7 x 5
= -1260

(ii) Let a = 28, b = -15 and c = -13
a + b + c = 28 + (-15) + (-13)
= 28 – 28 = 0
As a + b + c = 0
:.a3 + b3 + c3 = 3abc
= (28)3 + (-15)3 + (-13)3
= 3 x 28 x (-15) x (-13)
= 16380

Question 15.
Give possible expressions for the length and breadth of each of the following rectangles in which their areas are given:
(i) Area: 25a2 – 35a + 12
(ii) Area: 35y2 + 13y – 12
Solution:
(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
= (5a – 4)(5a – 3)
∴ The possible expressions for the length and breadth of the rectangle are 5a – 3 and 5a – 4.

(ii) 35y2 + 13y – 12
= 35y2 + 28y – 15y – l2
= 7y (5y + 4) – 3(5y + 4)
= (5y + 4)(7y – 3)
∴ The possible expressions for the length and breadth of the rectangle are 7y – 3 and 5y + 4.

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: 3x2 – 12x
(ii) Volume: 12ky2 + 8ky – 20k
Solution:
(i) 3x2 – 12x = 3x (x – 4)
∴ The possible expressions for the dimensions of the cuboid are 3, x and x – 4.

(ii) 12ky2 + 8ky – 20k
= 4k (3y2 + 2y – 5)
= 4k(3y2 + 5y – 3y – 5)
= 4k[y(3y + 5) – 1(3y + 5)]
= 4k(3y + 5)(y – 1)
∴ The possible expressions for the dimensions of the cuboid are 4k, 3y + 5 and y – 1.