PSEB Solutions for Class 9 Maths Chapter 2 Polynomials

 PSEB Solutions for Class 9 Maths Chapter 2 Polynomials

PSEB 9th Class Maths Solutions 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
Answer:
4x2 – 3x + 7 is a polynomial in one variable, because in each term the exponent of variable x is a whole number.

(ii) y2 + √2
Answer:
y2 + √2 is a polynomial in one variable, because in each term the exponent of variable y is a whole number.

(iii) 3√t + t√2
Answer:

(v) x10 + y3 + t50
Answer:
x10 + y3 + t50 is a polynomial, because the exponent of each of the variables x, y and t is a whole number, but it is not a polynomial in one variable.

Question 2.
Write the coefficients of x2 in each of the following:
(i) 2 + x2 + x
Answer:
In polynomial 2 + x2 + x, the coefficient of x2 is 1.

(ii) 2 – x2 + x3
Answer:
In polynomial 2 – x2 + x3, the coefficient of x2 is – 1.

(iv) √2x – 1
Answer:
In polynomial √2x – 1, the coefficient of x2 is 0 as the term with x2 is missing.

Question 3.
Give one example each of a binomial of degree 35 and of a monomial of degree 100.
Answer:
4x35 – 2, √5 x35 + 11, 7x35 – √3 x are a few examples of a binomial of degree 35. Similarly, 7x100, √2 x100, 4π x100 are a few examples of a monomial of degree 100.

Question 4.
Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x
Answer:
The degree of polynomial 5x3 + 4x2 + 7x is 3.

(ii) 4 – y2
Answer:
The degree of polynomial 4 – y2 is 2.

(iii) 5t – √7
Answer:
The degree of polynomial 5t – √7 is 1.

(iv) 3
Answer:
As the degree of constant polynomial 3 = 3x0 is 0.

Question 5.
Classify the following as and cubic polynomials:
(i) x2 + x
Answer:
As the degree of polynomial x2 + x is 2, it is a quadratic polynomial.

(ii) x – x3
Answer:
As the degree of polynomial x – x3 is 3 it is a cubic polynomial.

(iii) y + y2 + 4
Answer:
As the degree of polynomial y + y2 + 4 2. it is a quadratic polynomial.

(iv) 1 + x
Answer:
As the degree of polynomial 1 + x is 1, it is a linear polynomial.

(v) 3t
Answer:
As the degree of polynomial 3t is 1, it is a linear polynomial.

(vi) r2
Answer:
As the degree of polynomial r2 is 2, it is a quadratic polynomial.

(vii) 7x3
Answer:
As the degree of polynomial 7x3 is 3, it is a cubic polynomial.

Ex 2.2

Question 1.
Find the value of the polynomial 5x – 4x2 + 3 at (i) x = 0, (ii) x = – 1 and (iii) x = 2.
Answer:
Here, p (x) 5x – 4x2 + 3
(i) The value of polynomial p (x) at x = 0 is given by
p(0) = 5(0) – 4(0)2 + 3
= 3

(ii) The value of polynomial p (x) at x = – 1 is given by
p(- 1) = 5(- 1) – 4 (- 1)2 + 3
= – 5 – 4 + 3
= – 6

(iii) The value of polynomial p (x) at x = 2 is given by
p(2) = 5(2) – 4(2)2 + 3
= 10 – 16 + 3
= – 3

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p (y) = y2 – y + 1
Answer:
p(y) = y2 – y + 1
∴ p(0) = (0)2 – (0) + 1 = 1
∴ P(1) = (1)2 – (1) + 1 = 1 – 1 + 1 = 1
∴ p(2) = (2)2 – (2) + 1 = 4 – 2 + 1 = 3

(ii) p (t) = 2 + t + 2t2 – t3
Answer:
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 – (2)3
= 2 + 2 + 8 – 8 = 4

(iii) p(x) = x3
Answer:
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)
Answer:
p(x) = (x – 1) (x + 1)
∴ p(0) = (0 – 1) (0 + 1) = (- 1) × 1 = – 1
∴ p(1) = (1 – 1) (1 + 1) = 0 × 2 = 0
∴ p(2) = (2 – 1)(2 + 1) = 1 × 3 = 3

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

(iii) p(x) = x2 – 1, x = 1, – 1
Answer:
Here, p(x) = x2 – 1
Then, p(1) = (1)2 – 1 = 1 – 1 = 0 and
p(- 1) = (- 1)2 – 1 = 1 – 1 = 0.
Hence, 1 and – 1 both are zeroes of polynomial p(x) = x2 – 1.

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
Answer:
Here, p (x) = (x + 1) (x – 2)
Then, p(- 1) = (- 1 + 1) (- 1 – 2) = 0 × (-3)= 0
and p (2) = (2 + 1) (2 – 2) = 3 × O = O.
Hence, – 1 and 2 both are zeros of polynomial p(x) = (x + 1) (x – 2).

(v) p(x) = x2, x = 0
Answer:
Here, p(x) = x2
Then, p(0) = (0)2 = 0
Hence, 0 is a zero of polynomial p (x) = x2.

 

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Answer:
To find the zero of polynomial p (x) = x + 5,
we solve the equation p (x) = 0.
∴ x + 5 = 0
∴ x = – 5
Thus, – 5 is the zero of polynomial
p(x) = x + 5.

(ii) p(x) = x – 5
Answer:
To find the zero of polynomial p(x) = x – 5,
we solve the equation p (x) = 0.
∴ x – 5 = 0
∴ x = 5
Thus, 5 is the zero of polynomial
p(x) = x – 5.

(v) p(x) = 3x
Answer:
To find the zero of polynomial p (x) = 3x.
we solve the equation p (x) = 0.
∴ 3x = 0
∴ x = 0
Thus, 0 is the zero of polynomial P(x) = 3x.

(vi) p(x) = ax, a ≠ 0
Answer:
To find the zero of polynomial p (x) = ax,
a ≠ 0, we solve the equation p (x) = 0.
∴ ax = 0
∴ x = 0 (∵ a ≠ 0)
Thus, 0 is the zero of polynomial
p(x) = ax, a ≠ 0.

(vii) p(x) = cx + d, c ≠ 0, C, d are real numbers.
Answer:
To find the zero of polynomial
p(x) = cx + d, c ≠ 0, c, d are real numbers, we solve the equation p (x) = 0.

Ex 2.3

Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
Answer:
The remainder theorem states that when polynomial p (x) of degree greater than or equal to 1 is divided by linear polynomial x – a, the remainder is p (a).
Here. p(x) = x3 + 3x2 + 3x + 1

(i) x + 1
Answer:
Divisor g (x) = x + 1.
Comparing x + 1 with zero, we get x = – 1.
Then, remainder
= p(- 1)
= (- 1)3 + 3(- 1)2 + 3(- 1) + 1
= – 1 + 3 – 3 + 1
= 0

(iii) x
Answer:
Divisor g (x) = x.
x = 0 gives x = 0.
Then, remainder = p (0)
= (0)3 + 3(0)2 + 3(0) + 1
= 0 + 0 + 0 + 1
= 1

(iv) x + π
Answer:
Divisor g (x) = x + π.
x + π = 0 gives x = – π.
Then, remainder
= p(- π)
= (- π)3 + 3(- π)2 + 3(- π) + 1
= – π3 + 3π2 – 3π + 1

Question 2.
Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Answer:
Here, p (x) = x3 – ax2 + 6x – a and divisor
g (x) = x – a.
x – a = 0 gives x = a.
Then, remainder = p (a)
= (a)3 – a(a)2 + 6(a) – a
= a3 – a3 + 6a – a
= 5a

Question 3.
Check whether 7 + 3x is a factor of 3x3 + 7x.
Answer:
Here, p (x) = 3x3 + 7x and divisor g (x) = 7 + 3x.

Ex 2.4

Question 1.
Determine which of the following polynomials has (x + 1) a factor:
(i) x3 + x2 + x + 1
Answer:
The zero of x + 1 is – 1.
Let, p(x) = x3 + x2 + x + 1.
Then,
p(- 1) = (- 1)4 + (- 1)3 + (- 1)2 + (- 1) + 1
= – 1 + 1 – 1 + 1
≠ 0
So, by the factor theorem, x + 1 is a
factor of x4 + x3 + x2 + x + 1.

(ii) x3+ x3 + x2 + x + 1
Answer:
The zero of x +1 is – 1.
Let, p(x) = x4 + x3 + x2 + x + 1.
Then,
p(- 1) = (- 1)4 + (- 1)3 + (- 1)2 + (- 1) + 1
= 1 – 1 + 1 – 1 + 1
= 1
≠ 0
So, by the factor theorem, x + 1 is not a factor of x4 + x3 + x2 + x + 1.

(iii) x4 + 3x3 + 3x2 + x + 1
Answer:
The zero of x + 1 Is – 1.
Let, p(x )= x4 + 3x3 + 3x2 + x + 1.
Then,
p (-1) = (- 1)4 + 3 (- 1)3 + 3 (- 1)2 + (- 1) + 1
= 1 – 3 + 3 – 1 + 1
= 1
≠ 0
So, by the factor theorem, x + 1 is not a factor of x4 + 3x3 + 3x2 + x + 1.

(iv) x3 – x2 – (2 + √2)x + √2
Answer:
The zero of x + 1 is – 1. .
Let, p(x) = x3 – x2 – (2 + √2)x + √2.
Then.
p(- 1) = (- 1)3 – (- 1)2 – (2 + √2) (- 1) + √2
= – 1 – 1 + 2 + √2 + √2
= 2√2
≠ 0
So, by the factor theorem, x + 1 is not a factor of x3 – x2 – (2 + √2)x + √2.

Question 2.
Use the factor theorem to determine whether g (x) is a factor of p (x) in each of the following cases:
(i) p (x) = 2x3 + x2 – 2x – 1, g (x) = x + 1
Answer:
g(x) = 0 gives x + 1 = 0, i.e., x = – 1.
Here, p(x) = 2x3 + x2 – 2x – 1
Then, p(- 1) = 2(- 1)3 + (- 1)2 – 2(- 1) – 1
= – 2 + 1 + 2 – 1
= 0
So, by the factor theorem, g (x) = x + 1 is a factor of p(x) = 2x3 + x2 – 2x – 1.

(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
Answer:
g(x) = 0 gives x + 2 = 0, i.e., x = – 2.
Here, p(x) = x3 + 3x2 + 3x + 1
Then, p(- 2) = (- 2)3 + 3(2)2 + 3(- 2) + 1
= – 8 + 12 – 6 + 1
≠ 0
So, by the factor theorem, g (x) = x + 2 is not a factor of p(x) = x3 + 3x2 + 3x + 1.

(iii) p (x) = x3 – 4x2 + x + 6. g (x) = x – 3
Answer:
g(x) = 0 gives x – 3 = 0, i.e., x = 3.
Here, p (x) = x3 – 4x2 + x + 6
Then, p(3)= (3)3 – 4(3)2 + (3) + 6
= 27 – 36 + 3 + 6
= 0
So, by the factor theorem, g (x) = x – 3 is a factor of p (x) = x3 – 4x2 + x + 6.

Question 3.
Find the value of k, If x – 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x2 + x + k
Answer:
Here x – 1 is a factor of p(x) = x2 + x + k.
∴ p(1) = 0
∴ (1)2 + (1) + k = 0
∴ 1 + 1 + k = 0
∴ 2 + k = 0
∴ k = – 2

(ii) p (x) = 2x2 + kx + √2
Answer:
Here, x – 1 is a factor of
p(x) = 2x2 + kx + √2.
∴ p(1) = 0
∴ 2(1)2 + k(1) + √2 = 0
∴ 2 + k + √2 = 0
∴ k = -(2 + √2)

(iii) p (x) = kx2 – √2x + 1
Answer:
Here, x – 1 is a factor of
p(x) = kx2 – √2x + 1.
∴ p (1) = 0
∴ k(1)2 – √2 (1) + 1 = 0
∴ k – √2 + 1 = 0
∴ k = √2 – 1

(iv) p (x) = kx2 – 3x + k
Answer:
Here, x – 1 is a factor of p (x) = kx2 – 3x + k.
∴ p(1) = 0
∴ k(1)2 – 3(1) + k = 0
∴ k – 3 + k = 0
∴ 2k = 3
∴ k = [1atex]\frac{3}{2}[/1atex]

Question 4.
Factorise:
(i) 12x2 – 7x + 1
Answer:
12x2 – 7x + 1 = 12x2 – 4x – 3x + 1
= 4x(3x – 1) – 1(3x – 1)
= (3x – 1)(4x – 1)

(ii) 2x2 + 7x + 3
Answer:
2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3) (2x + 1)

(iii) 6x2 + 5x – 6
Answer:
6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)

(iv) 3x2 – x – 4
Answer:
3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1) (3x – 4)

Question 5.
Factorise:
(i) x3 – 2x2 – x + 2
Answer:
Let, p(x) = x3 – 2x2 – x + 2
All the factors of 2 are ± 1 and ± 2.
By trial, we find that p (1) = 0.
∴ x – 1 is a factor of p (x).
x3 – 2x2 – x + 2
= x3 – x2 – x2 + x – 2x + 2
= x2(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)

(ii) x3 – 3x2 – 9x – 5
Answer:
Let, p(x) = x3 – 3x2 – 9x – 5
All the factors of – 5 are ± 1 and ±5.
By trial, we find that p (- 1) = 0.
∴ x + 1 is a factor of p (x).
x3 – 3x2 – 9x – 5
= x3 + x2 – 4x2 – 4x – 5x – 5
= x2(x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1) (x2 – 4x – 5)
= (x + 1) (x2 + x – 5x – 5)
= (x + 1) {x(x + 1) – 5(x + 1)}
= (x + 1)(x + 1)(x – 5)

(iii) x3 + 13x2 + 32x + 20
Answer:
Let, p (x) = x3 + 13x2 + 32x + 20
All the factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10 and ± 20.
By trial, we find that p (- 1) = 0.
∴ x + 1 is a factor of p (x).
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 + 2x + 10x + 20)
= (x + 1) {x(x + 2) + 10(x + 2)}
= (x + 1) (x + 2) (x + 10)

(iv) 2y3 + y2 – 2y – 1
Answer:
Let, p (y) = 2y3 + y2 – 2y – 1
All the factors of – 1 are ± 1.
By trial, we find that p (- 1) = 0.
∴ y + 1 is a factor of p (y).
2y3 + y2 – 2y – 1
= 2y3 + 2y2 – y2 – y – y – 1
= 2y2(y + 1) – y(y + 1) – 1 (y + 1)
= (y + 1) (2y2 – y – 1)
= (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)
Answer:
(x + 4) (x + 10)
= (x)2 + (4 + 10)x + (4)(10)
= x2 + 14x + 40

(ii) (x + 8) (x – 10)
Answer:
(x + 8) (x – 10)
= (x)2 + (8 – 10)x + (8)(- 10)
= x2 – 2x – 80

(iii) (3x + 4) (3x – 5)
Answer:
(3x + 4) (3x – 5)
= (3x)2 + (4 – 5) (3x) +(4) (- 5)
= 9x2 – 3x – 20

(v)(3 – 2x) (3 + 2x)
Answer:
(3 – 2x) (3 + 2x)
= (3)2 – (2x)2
= 9 – 4x2

Question 2.
Evaluate the following products without multiplying directly:
(i) 103 × 107
Answer:
103 × 107
= (100 + 3) (100 + 7)
= (100)2 + (3 + 7) (100) + (3) (7)
= 10000 + 1000 + 21
= 11,021

(ii) 95 × 96
Answer:
95 × 96
= (90 + 5) (90 + 6)
= (90)2 + (5 + 6) (90) + (5) (6)
= 8100 + 990 + 30 = 9120

OR

95 × 96
= (100 – 5) (100 – 4)
= (100)2 + (- 5 – 4) (100) + (- 5) (- 4)
= 10000 – 900 + 20
= 10020 – 900 = 9120

(iii) 104 × 96
Answer:
104 × 96
= (100 + 4)(100-4)
= (100)2 – (4)2
= 10000 – 16 = 9984

Question 3.
Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2
Answer:
9x2 + 6xy + y2
= (3x)2 + 2(3x) (y) + (y)2
= (3x + y)2 = (3x + y) (3x + y)

(ii) 4y2 – 4y + 1
Answer:
4y2 – 4y + 1
= (2y)2 – 2(2y)(1) + (1)2
= (2y – 1)2 = (2y – 1) (2y – 1)

Question 4.
Expand each of the following using suitable identities:
(i) (x + 2y + 4z)2
Answer:
(x + 2y + 4z)2
= (x)2 + (2y)2 + (4z)2 + 2 (x) (2y) + 2(2y)(4z) + 2(4z)(x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

(ii) (2x – y + z)2
Answer:
(2x – y + z)2
= [2x + (- y) + z]2
= (2x)2 + (- y)2 + (z)2 + 2 (2x)(- y) + 2(- y)(z) + 2(z) (2x)
= 4x2 + y2 + z2 – 4xy – 2yz + 4zx

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

(iv) (3a – 7b – c)2
Answer:
(3a – 7b – c)2
= [3a + (- 7b) + (- c)]2
= (3a)2 + (- 7b)2 + (- c)2 + 2 (3a) (- 7b) + 2(- 7b) (- c) + 2(- c) (3a)
= 9a2 + 49b2 + c2 – 42ab + 14bc – 6ca

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

Question 5.
Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
Answer:
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz ;
= (2x)2 + (3y)2 + (- 4z)2 + 2 (2x) (3y) + 2(3y) (- 4z) + 2(- 4z) (2x)
= [2x + 3y + (- 4z)]2
= (2x + 3y – 4z)2
= (2x + 3y – 4z) (2x + 3y – 4z)

(ii) 2x2 + y2 + 8z2 – 2√2 xy + 4√2 yz – 8xz
Answer:
2x2 + y2 + 8z2 – 2√2 xy + 4√2 yz – 8 xz
= (- √2x)2 + (y)2 + (2√2 z)2 + 2 (- √2x) (y) + 2(y) (2√2 z) + 2(2√2 z) (- √2 x)
= [(- √2 x) + y + 2√2 z]2
= (- √2x + y + 2√2 z)2
= (- √2x + y + 2√2 z) (- √2 x + y + 2√2 z)

Question 6.
Write the following cubes in expanded form:
(i) (2x + 1)3
Answer:
(2x + 1)3
= (2x)3 + (1)3 + 3(2x) (1)(2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3+ 1 + 12x2 + 6x
OR
(2x + 1)3
= (2x)3 + 3(2x)2(1) + 3(2x) (1)2 + (1)3
= 8x3 + 12x2 + 6x + 1

(ii) (2a – 3b)3
Answer:
(2a – 3b)3
= (2a)3 – (3b)3 – 3 (2a) (3b) (2a – 3b)
= 8a3 – 27b3 – 18ab(2a – 3b)
= 8a3 – 27b3 – 36a2b + 54ab2

Question 7.
Evaluate the following using suitable identities:
(i) (99)3
Answer:
(99)3 = (100 – 1)3
= (100)3 – (1)3 – 3 (100) (1) (100 – 1)
= 1000000 – 1 – 300(99)
= 1000000 – 1 – 29700
= 9,70,299

(ii) (102)3
Answer:
(102)3 = (100 + 2)3
= (100)3 + (2)3 + 3(100) (2) (100 + 2)
= 1000000 + 8 + 600(102)
= 1000000 + 8 + 61200
= 10,61,208

(iii) (998)3
Answer:
(998)3 = (1000 – 2)3
= (1000)3 – (2)3 – 3(1000) (2) (1000 – 2)
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 + 12000
= 994000000 + 12000 – 8
= 994012000 – 8
= 99,40,11,992

Question 8.
Factorise each of the following:
(i) 8a3 + b3 + 12a2b + 6ab2
Answer:
8a3 + b3 + 12a2b + 6ab2
= (2a)3 + (b)3 + 3 (4a2) (b) + 3 (2a) (b2)
= (2a)3 + (b)3 + 3 (2a)2 (b) + 3 (2a) (b2)
= (2a + b)3
= (2a + b) (2a + b) (2a + b)

(ii) 8a3 – b3 – 12a2b + 6ab2
Answer:
8a3 – b3 – 12a2b + 6ab2
= (2a)3 + (- b)3 + 3 (4a2) (- b) + 3 (2a) (b2)
= (2a)3 + (- b)3 + 3 (2a)3 (- b) + 3 (2a) (- b)2
= (2a – b)3
= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a3 – 135a + 225a2
Answer:
27 – 125a3 – 135a + 225a2
= (3)3 + (- 5a)3 + 3(9) (- 5a) + 3 (3) (25a2)
= (3)3 + (- 5a)3 + 3(3)2 (- 5a) + 3 (3) (- 5a)2
=(3 – 5a)3.
= (3 – 5a)(3 – 5a) (3 – 5a)

(iv) 64a3 – 27b3 – 144a2b + 108ab2
Answer:
64a3 – 27b3 – 144a2b + 108ab2
= (4a)3 + (- 3b)3 + 3(16a2)(-3b) + 3(4a) (9b2)
= (4a)3 + (- 3b)3 + 3(4a)2(- 3b) + 3(4a)(- 3b)2
= (4a – 3b)3
= (4a – 3b) (4a – 3b) (4a – 3b)

Question 9.
Verify:
(i) x3 + y3 = (x + y) (x2 – xy + y2)
Answer:
R.H.S. = (x + y) (x2 – xy + y2)
= x(x2 – xy + y2) + y(x2 – xy + y2)
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 + y3
= L.H.S.

(ii) x3 – y3 = (x – y) (x2 + xy + y2)
Answer:
R.H.S. = (x – y) (x2 + xy + y2)
= x(x2 + xy + y2) – y(x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3
= L.H.S.

Question 10.
Factorise each of the following:
[Hint: See Question 9]
(i) 27y3 + 125z3
Answer:
27y3 + 125z3
We know, a3 + b3 = (a + b) (a2 – ab + b2)
Replacing a by 3y and b by 5z, we get
(3y)3 + (5z)3 = (3y + 5z) [(3y)2 – (3y)(5z) + (5z)2]
∴ 27y3 + 125z3 = (3y + 5z) (9y2 – 15yz + 25z2)

OR

We know, a3 + b3 = (a + b)3 – 3ab(a + b)
Replacing a by 3y and b by 5z, we get
(3y)3 + (5z)3 = (3y + 5z)3 – 3 (3y) (5z) (3y + 5z)
∴ 27y3 + 125z3 = (3y + 5z) [(3y + 5z)3 – 45yz]
= (3y + 5z)(9y3 + 30yz + 25z2 – 45yz)
= (3y + 5z) (9y2 – 15yz + 25z2)

(ii) 64m3 – 343n3
Answer:
We know, a3 – b3 = (a – b) (a2 + ab + b2)
Replacing a by 4m and b by 7n, we get
(4m)3 – (7n)3 = (4m – 7n) [(4m)3 + (4m) (7n) + (7n)2]
∴ 64m3 – 343n3 = (4m – 7n) (16m2 + 28mn + 49n2)

OR

We know. a3 – b3 = (a – b)3 + 3ab(a – b)
Replacing a by 4m and b by 7n, we get
(4m)3 – (7n)3 = (4m – 7n)3 + 3 (4m) (7n)(4m – 7n)
= (4m – 7n) [(4m 7n)2 + 84mn]
= (4m – 7n) (16m2 – 56mn + 49n2 + 84mn)
= (4m – 7n) (16m2 + 28mn + 49n2)

Question 11.
Factorise: 27x3 + y3 + z3 – 9xyz
Answer:
We know, a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
Replacing a by 3x, b by y and c by z, we get
(3x)3 + (y)3 + (z)3 – 3 (3x) (y) (z) = (3x + y + z) [(3x)2 + (y)2 + (z)2– (3x) (y) – (y) (z) — (z) (3x)]
∴ 27x3 + y3 + z3 – 9xyz = (3x + y + z) (9x2 + y2 + z2– 3xy – yz – 3zx)

Question 12.
Verify that x3 + y3 + z3 – 3xyz = 1/2(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2)
Answer:
R.H.S. = 1/2(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
= 1/2 (x + y + z) (x2 – 2xy + y2 + y2 – 2yz + z2 + z2 – 2zx + x2)
= 1/2 (x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= x(x2 + y2 + z2 – xy – yz – zx) + y(x2 + y2 + z2 – xy – yz – zx) + z(x2 + y2 + z2 – xy – yz – zx)
= x3 + xy2 + xz2 – x2y – xyz – zx2 + x2y + y3 + yz2 – xy2 – y2z – xyz + x2z + y2z + z3 – xyz – yz2 – z2x
= x3 + y3 + z3 – 3xyz
= L.H.S.

Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Answer:
We know the Idendity
x3 + y3 + z3 – 3xyz = (x + y + z) (x2+ y3 + z2 – xy – yz – zx)
If x + y + z = 0. we get
x3 + y3 + z3 – 3xyz = (0) (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
Answer:
Taking a = -12, b = 7 and c = 5, we get
a + b + c = (- 12) + 7 + 50.
Now, If a + b + c = 0, then a3 + b3 + c3 = 3abc.
∴ (- 12)3 + (7)3 + (5)3 = 3(- 12) (7) (5)
= (- 36) (35)
= – 1260

(ii) (28)3 + (- 15)3 + (- 13)3
Answer:
Talking a = 28, b = – 15 and c = – 13, we get
a + b + c = 28 + (- 15) + (- 13) = 0.
Now, If a + b + c = 0, then a3 + b3 + c3 = 3abc.
∴ (28)3 + (- 15)3 + (- 13)3 = 3 (28) (- 15) (- 13)
= (84)(195)
= 16,380

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
Answer:
We know, area of a rectangle = length × breadth
Hence, two factors of area can give possible expressions for length and breadth. So here, we will try to obtain two factors of the expression of area.
25a2 – 35a + 12 = 25a2 – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
=(5a – 4) (5a – 3)
Thus, the length and breadth of the rectangle are (5a – 3) and (5a – 4) respectively.
Note : Traditionally, length > breadth in a rectangle.

(ii) Area: 35y2 + 13y – 12
Answer:
We know, area of a rectangle = length × breadth
Hence, two factors of area can give possible expressions for length and breadth. So here, we will try to obtain two factors of the expression of area.
35y2 + 13y – 12 = 35y2 + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
= (5y + 4) (7y – 3)
Thus, the length and breadth of the rectangle are (7y – 3) and (5y + 4) respectively.

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
Answer:
(i) Volume: 3x2 – 12x
Answer:
We know, volume of a cuboid = length × breadth × height
Hence, three factors of volume can give possible expressions for length, breadth and height.
So here, we will try to obtain three factors of the expression of volume.
3x2 – 12x = 3x(x – 4)
= 3 × x × (x – 4)
Thus, one possible answer for the dimensions of the cuboid is 3. x and (x – 4).
Note: Other possible answers can be given as 1. 3x and (x – 4) or 1, x and (3x – 12).

(ii) Volume: 12ky2 + 8ky = 20k
Answer:
We know, volume of a cuboid = length × breadth × height
Hence, three factors of volume can give possible expressions for length, breadth and height.
So here, we will try to obtain three factors of the expression of volume.
12ky2 + 8ky – 20k = 4k (3y2 + 2y – 5)
= 4k (3y2 – 3y + 5y – 5)
= 4k [3y (y – 1) + 5(y – 1)1]
= 4k (y – 1) (3y + 5)
Thus, one possible answer for the dimensions of the cuboid is 4k, (y – 1) and (3y + 5).

MCQ

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The value of p(x) = x3 + x2 – 3x – 3 at x = – 1 is ………….. .
A. 1
B. – 1
C. 0
D. – 3
Answer:
C. 0

Question 2.
For the polynomial p(x), If p(2) = 0. then …………….. is a factor of p(x).
A. (x – 2)
B. (x + 2)
C. (x2 – 2)
D. (x2 + 2)
Answer:
A. (x – 2)

Question 3.
Dividing 2x3 + 6x2 + x + 5 by (x + 3), the remainder is ………………. .
A. 2
B. 3
C. 1
D. 0
Answer:
A. 2

Question 4.
……………. should be subtracted from x3 + 3x2 + 2x + 10, so that the result is exactly divisible by (x + 3).
A. 1
B. 2
C. 3
D. 4
Answer:
D. 4

Question 5.
………………… should be added to x3 – 5x2 + x – 8, so that the result Is exactly divisible by (x – 5).
A. 2
B. – 2
C. – 3
D. 3
Answer:
D. 3

Question 6.
…………………. is one of the zeros of the polynomial x3 – 6x2 + 2x – 12.
A. 2
B. – 2
C. 6
D. – 6
Answer:
C. 6

Question 7.
If x + 3 is a factor of x3 + 6x2 + 11x + k, then k = ……………….. .
A. 2
B. 3
C. 4
D. 6
Answer:
D. 6

Question 8.
………………….. should be added to x2 – 8, so that the result is exactly divisible by (x + 3).
A. 1
B. – 1
C. 3
D. – 3
Answer:
B. – 1

Question 9.
If x – 1 is a factor of 4x3 + 3x2 – 4x + k, then k = ………… .
A. 4
B. 1
C. 3
D. – 3
Answer:
D. – 3

Question 10
………………. is one of the factors of 2x4 + x3 – 14x2 – 19x – 6.
A. (x – 1)
B. (x + 1)
C. (x + 3)
D. (x – 2)
Answer:
B. (x + 1)

Question 11.
When x3 + 64 is divided by x + 4, the quotient is ………………. .
A. (x – 4)
B. (x + 8)
C. (x + 16)
D. x2 – 4x + 16
Answer:
D. x2 – 4x + 16

Question 12.
x3 + 3x2 + 3x + 2 = (x + 2) (……………..)
A. x – 2
B. x2 + 1
C. x2 – x – 1
D. x2 + x + 1
Answer:
D. x2 + x + 1

Question 13.
The factors of x2 – x – 12 are .
A.(x + 6) and (x – 2)
B. (x – 4) and (x – 3)
C. (x + 4) and (x + 3)
D. (x – 4) and (x + 3)
Answer:
D. (x – 4) and (x + 3)

Question 14.
The factors of x2 – 100 are ………………. .
A. (x – 20) and (x – 5)
B. (x – 25) and (x – 4)
C. (x – 10)2
D. (x + 10) and (x – 10)
Answer:
D. (x + 10) and (x – 10)

Question 16.
If x2 + mx – 28 = (x – 7) (x + 4), then m = ………………….. .
A. 3
B. – 3
C. 11
D. – 11
Answer:
B. – 3

Question 17.
105 × 95 = ………………. .
A. 9500
B. 10,500
C. 9925
D. 9975
Answer:
D. 9975

Question 18.
(110)3 = …………………… .
A. 330
B. 3300
C. 33.000
D. 13,31,000
Answer:
D. 13,31,000

Question 19.
(15)3 – (9)3 – (6)3 = ………………….
A. 1215
B. 2430
C. – 810
D. 810
Answer:
B. 2430

Question 20.
If x + 3 is one of the factors x3 + 2x2 – ax – 18, then a = …………………. .
A. 3
B. – 3
C. 9
D. – 9
Answer:
C. 9