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

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) x3 + x2 + x + 1
(ii) x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x +1
(iv) x3 – x2 – (2 + √2)x + √2
Solution:
(i) Here, p(x) = x3 + x2 + 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 x3 + x2 + x + 1.

(ii) Here, p(x) = x4 + x3 + x2 + 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 x4 + x3 + x2 + x + 1.

(iii) Here, p(x) = x4 + 3x3 + 3x2 + 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 x4 + 3x3 + 3x2 + x + 1.

(iv) Here, p(x) = x3 – x2 – (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 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
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
(iii) pix) = x3 – 4x2 + x + 6, g(x) = x – 3
Solution:
(i) Here, p(x) = 2x3 + x2 – 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) = 2x3 + x2 – 2x – 1

(ii) Here, p(x) = x3 + 3x2 + 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) = x3 + 3x2 + 3x + 1

(iii) Here, p(x) = x3 – 4x2 + 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) = 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
(ii) p(x) = 2x2 + kx + √2
(iii) p(x) = kx2 – √2x + 1
(iv) p(x) = kx2 – 3x + k
Solution:
(i) We have give that x – 1 is the factor of p(x) = x2 + 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) = x2 + x + k is -2.

(ii) We have given that x – 1 is the factor of P(x) = 2x2 + 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) = 2x2 + kx + 2 is -(2 + √2)

(iii) We have given that x – 1 is a factor of
p(x) = kx2 – √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) = kx2 – √2x + 1 is √2 – 1.

(iv) We have given that x – 1 is the factor of p(x) = kx2 – 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) = kx2 – 3x + k is $$\frac{3}{2}$$

Question 4.
Factorise
(i) 12x2 – 7x + 1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4
Solution:
(i) we have given 12x2 – 7x + 1
We can also write 12x2 – 4x – 3x – 1
or, 4x(3x – 1) – 1(3x – 1)
or, (3x – 1)(4x – 1)

(ii) We have given 2x2 + 7x + 3
We can also write 2x2 + 6x + x + 3
or, 2x(x + 3) + 1(x + 3)
or, (x + 3) (2x + 1)

(iii) We have given 6x2 + 5x – 6
We can also write 6x2 + 9x – 4x – 6
or, 3x(2x + 3) – 2(2x + 3)
or, (2x + 3) (3x – 2)

(iv) We have given 3x2 – x – 4
We can also write 3x2 – 4x + 3x – 4
or, x(3x – 4) + 1(3x – 4)
or, (3x – 4)(x + 1)

Question 5.
Factorise:
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 – 13x2 + 20
(iv) 2y3 + y2 – 2y – 1
Solution:
(i) Let p(x) = x3 – 2x2 – 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, x3 – 2x2 – x + 2
= (x – 2)(x2 – 1)
= (x – 2)(x + 1)(x – 1)
[∵ a2 – b2 = (a + b)(a – b)]
∴ x3 – 2x2 – x + 2 = (x – 2)(x – 1)(x + 1)

(ii) Let P(x) = x3 – 3x2 – 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,
x3 – 3x2 – 9x – 5
= (x + 1) (x2 – 4x – 5)
= (x + 1) [x2 – 5x + x – 5]
= (x + 1) [x(x – 5) + 1(x – 5)]
∴ x3 – 3x2 – 9x – 5 = (x + 1)(x + 1)(x – 5)

(iii) Let P(x) = x3 + 13x2 + 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,
x3 + 12x2 + 32x + 20
= (x + 1) (x2 + 12x + 20)
= (x + 1) [x2 + 10x + 2x + 20]
= (x + 1) [x(x +10) + 2(x + 10)]
∴ x3 + 13x2 + 32x + 20 = (x + 1)(x + 10) (x + 2)

(iv) Let P(y) = 2y3 + y2 – 2y – 1
The factor of -1 are ±1
By trial, we find that P(1) = 0
So, (y – 1) must be the factor of 2y3 + y2 – 2y – 1
Therefore,
2y3 + y2 – 2y – 1
= (y – 1) (2y2 + 3y + 1)
= (y – 1) [2y2 + 2y + y + 1]
= (y – 1) [2y(y + 1) + 1(y + 1)]
∴ 2y3 + y2 – 2y – 1 = (y – 1)(y + 1)(2y + 1)

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