These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4

Question 1.

Which of the following polynomials has (x + 1) a factor:

(i) x^{3} + x^{2} + x + 1

(ii) x^{4} + x^{3} + x^{2} + x + 1

(iii) x^{4} + 3x^{3} + 3x^{2} + x +1

(iv) x^{3} – x^{2} – (2 + √2)x + √2

Solution:

(i) Here, p(x) = x^{3} + x^{2} + x + 1

and the zero of x + 1 is -1.

So, p(-1) = (-1)^{3} + (-1)^{2} + (-1) + 1

= -1 + 1 – 1 + 1

= 0

Here, remainder is zero, therefore x + 1 is the factor of x^{3} + x^{2} + x + 1.

(ii) Here, p(x) = x^{4} + x^{3} + x^{2} + x + 1

and the zero of x + 1 is -1.

So p(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1) + 1

= 1 – 1 + 1 – 1 + 1

= 1

Here, remainder is 1, therefore x + 1 is not the factor of x^{4} + x^{3} + x^{2} + x + 1.

(iii) Here, p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1

and the zero of x + 1 is -1

So, p(-1) = (-1)^{4} + 3(-1)^{3} + 3(-1)^{2} + (-1) + 1

= 1 – 3 + 3 – 1 + 1

= 1

Here, remainder is 1, therefore x + 1 is not the factor of x^{4} + 3x^{3} + 3x^{2} + x + 1.

(iv) Here, p(x) = x^{3} – x^{2} – (2 + √2)x + √2 and the zero of x + 1 is -1

So, p(-1) = (-1)^{3} – (-1)^{2} – (2 + √2) (-1) + √2

= -1 – 1 + 2 + √2 + √2

= 2√2

Here, remainder is 2√2 therefore x + 1 is not the factor of x^{3} – x^{2} – (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) = 2x^{3} + x^{2} – 2x – 1, g(x) = x + 1

(ii) p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

(iii) pix) = x^{3} – 4x^{2} + x + 6, g(x) = x – 3

Solution:

(i) Here, p(x) = 2x^{3} + x^{2} – 2x – 1 and the zero of g(x) = x + 1 is -1

So p(-1) = 2(-1)^{3} + (-1)^{2} – 2(-1) – 1

= -2 + 1 + 2 – 1

= 0

Here, remainder is zero, therefore g(x) = x + 1 is the factor of p(x) = 2x^{3} + x^{2} – 2x – 1

(ii) Here, p(x) = x^{3} + 3x^{2} + 3x + 1

and the zero of g(x) = x + 2 is -2.

So, p(-2) = (-2)^{3} + 3(-2)^{2} + 3(-2) + 1

= -8 + 12 – 6 + 1

= -1

Here, remainder is -1, therefore g(x) = x + 2 is not the factor of p(x) = x^{3} + 3x^{2} + 3x + 1

(iii) Here, p(x) = x^{3} – 4x^{2} + x + 6

and the zero of g(x) = x – 3 is 3.

So, p(3) = (3)^{3} – 4(3)^{2} + 3 + 6

= 27 – 36 + 3 + 6

= 0

Here, remainder is 0, therefore g(x) = x – 3 is the factor of p(x) = x^{3} – 4x^{2} + 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) = x^{2} + x + k

(ii) p(x) = 2x^{2} + kx + √2

(iii) p(x) = kx^{2} – √2x + 1

(iv) p(x) = kx^{2} – 3x + k

Solution:

(i) We have give that x – 1 is the factor of p(x) = x^{2} + x + k

So, (1)^{2} + 1 + k = 0

or, 1 + 1 + k = 0

2 + k = 0

k = -2

Therefore, the value of k in p(x) = x^{2} + x + k is -2.

(ii) We have given that x – 1 is the factor of P(x) = 2x^{2} + kx + √2

P(1) = 0

Now, P(1) = 2(1)^{2} + k(1) + √2

So, 0 = 2 + k + √2

or, k = -2 – √2

or, k = -(2 + √2)

Therefore the value of k in p(x) = 2x^{2} + kx + 2 is -(2 + √2)

(iii) We have given that x – 1 is a factor of

p(x) = kx^{2} – √2x + 1

p(1) = 0

Now, p(1) = k(1)^{2} – √2(1) + 1

So, k – √2 + 1 = 0

k = √2 – 1

Therefore, the value of it in p(x) = kx^{2} – √2x + 1 is √2 – 1.

(iv) We have given that x – 1 is the factor of p(x) = kx^{2} – 3x + k

P(1) = 0

Now, p(1) = k(1)^{2} – 3(1) + k

So, k – 3 + k = 0

or, 2k – 3 = 0

k = \(\frac{3}{2}\)

Therefore the value of k in p(x) = kx^{2} – 3x + k is \(\frac{3}{2}\)

Question 4.

Factorise

(i) 12x^{2} – 7x + 1

(ii) 2x^{2} + 7x + 3

(iii) 6x^{2} + 5x – 6

(iv) 3x^{2} – x – 4

Solution:

(i) we have given 12x^{2} – 7x + 1

We can also write 12x^{2} – 4x – 3x – 1

or, 4x(3x – 1) – 1(3x – 1)

or, (3x – 1)(4x – 1)

(ii) We have given 2x^{2} + 7x + 3

We can also write 2x^{2} + 6x + x + 3

or, 2x(x + 3) + 1(x + 3)

or, (x + 3) (2x + 1)

(iii) We have given 6x^{2} + 5x – 6

We can also write 6x^{2} + 9x – 4x – 6

or, 3x(2x + 3) – 2(2x + 3)

or, (2x + 3) (3x – 2)

(iv) We have given 3x^{2} – x – 4

We can also write 3x^{2} – 4x + 3x – 4

or, x(3x – 4) + 1(3x – 4)

or, (3x – 4)(x + 1)

Question 5.

Factorise:

(i) x^{3} – 2x^{2} – x + 2

(ii) x^{3} – 3x^{2} – 9x – 5

(iii) x^{3} – 13x^{2} + 20

(iv) 2y^{3} + y^{2} – 2y – 1

Solution:

(i) Let p(x) = x^{3} – 2x^{2} – x + 2

The factors of 2 are ±1, ±2

By trial, we find that p(2) = 0

So, (x – 2) must be the factor of p(x)

Therefore, x^{3} – 2x^{2} – x + 2

= (x – 2)(x^{2} – 1)

= (x – 2)(x + 1)(x – 1)

[∵ a^{2} – b^{2} = (a + b)(a – b)]

∴ x^{3} – 2x^{2} – x + 2 = (x – 2)(x – 1)(x + 1)

(ii) Let P(x) = x^{3} – 3x^{2} – 9x – 5

The factors of -5 are ±1, ±5

By trial, we find that P(-1) = 0

So, (x + 1) is the fact or of p(x)

Therefore,

x^{3} – 3x^{2} – 9x – 5

= (x + 1) (x^{2} – 4x – 5)

= (x + 1) [x^{2} – 5x + x – 5]

= (x + 1) [x(x – 5) + 1(x – 5)]

∴ x^{3} – 3x^{2} – 9x – 5 = (x + 1)(x + 1)(x – 5)

(iii) Let P(x) = x^{3} + 13x^{2} + 32x + 20

The factor of 20 are +1, ±2, ±4, ±5, ±10, ±20.

By trial, we find that P(-1) = 0

So, (x +1) must be the factor of P(x) Therefore,

x^{3} + 12x^{2} + 32x + 20

= (x + 1) (x^{2} + 12x + 20)

= (x + 1) [x^{2} + 10x + 2x + 20]

= (x + 1) [x(x +10) + 2(x + 10)]

∴ x^{3} + 13x^{2} + 32x + 20 = (x + 1)(x + 10) (x + 2)

(iv) Let P(y) = 2y^{3} + y^{2} – 2y – 1

The factor of -1 are ±1

By trial, we find that P(1) = 0

So, (y – 1) must be the factor of 2y^{3} + y^{2} – 2y – 1

Therefore,

2y^{3} + y^{2} – 2y – 1

= (y – 1) (2y^{2} + 3y + 1)

= (y – 1) [2y^{2} + 2y + y + 1]

= (y – 1) [2y(y + 1) + 1(y + 1)]

∴ 2y^{3} + y^{2} – 2y – 1 = (y – 1)(y + 1)(2y + 1)