NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.4

NCERT Solutions Class 9 Maths
Chapter – 2 (Polynomials)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 2 Polynomials Exercise 2.4 has been provided here to help the students in solving the questions from this exercise.

Chapter 2: Polynomials

Exercise – 2.4

1. Determine 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

Answer –
(i) x³+ x²+ x + 1
Let p(x) = x³+ x²+ x + 1
The zero of x + 1 is -1. [x + 1 = 0 means x = -1]
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.

(ii) x⁴+ x³+ x²+ x + 1
Let p(x) = x⁴+ x³+ x²+ x + 1
The zero of x + 1 is -1. [x + 1 = 0 means x = -1]
p(−1) = (−1)⁴+ (−1)³+ (−1)²+ (−1) + 1
= 1 − 1 + 1 − 1 + 1
= 1 ≠ 0
∴ By factor theorem, x + 1 is not a factor of x⁴ + x³ + x² + x + 1.

(iii) x⁴+ 3x³+ 3x²+ x + 1
Let p(x)= x⁴+ 3x³+ 3x²+ x + 1
The zero of x + 1 is -1.
p(−1) = (−1)⁴+ 3(−1)³+ 3(−1)²+ (−1) + 1
=1 − 3 + 3 − 1 + 1
=1 ≠ 0
∴ By factor theorem, x+1 is not a factor of x⁴+ 3x³+ 3x²+ x +1.

(iv) x³– x²– (2 + √2)x + √2
Let p(x) = x³– x²– (2 + √2)x + √2
The zero of x + 1 is -1.
p(−1) = (-1)³– (-1)²– (2 + √2)(-1) + √2
= −1 − 1 + 2 + √2 + √2
= 2√2 ≠ 0
∴ By factor theorem, x + 1 is not a factor of x³– x²– (2 + √2)x + √2.

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) p (x) = x³ – 4x² + x + 6, g (x) = x – 3

Answer –
(i) p(x) = 2x³+ x²– 2x – 1, g(x) = x + 1
p(x) = 2x³+ x²– 2x – 1, g(x) = x+1
g(x) = 0
⇒ x + 1 = 0
⇒ x = −1
∴ Zero of g(x) is -1.
Now,
p(−1) = 2(−1)³+ (−1)²– 2(−1) – 1
= −2 + 1 + 2 − 1
= 0
∴ By factor theorem, g(x) is a factor of p(x).

(ii) p(x) = x³+ 3x²+ 3x + 1, g(x) = x+2
g(x) = 0
⇒ x + 2 = 0
⇒ x = −2
∴ Zero of g(x) is -2.
Now,
p(−2) = (−2)³+ 3(−2)²+ 3(−2) + 1
= −8 + 12 − 6 + 1
= −1 ≠ 0
∴ By factor theorem, g(x) is not a factor of p(x).

(iii) p(x) = x³– 4x²+ x + 6, g(x) = x – 3
g(x) = 0
⇒ x − 3 = 0
⇒ x = 3
∴ Zero of g(x) is 3.
Now,
p(3) = (3)³− 4(3)²+ (3) + 6
= 27 − 36 + 3 + 6
= 0
∴ By factor theorem, g(x) is a factor of p(x).

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 – √2 x + 1
(iv) p (x) = kx² – 3x + k

Answer –
(i) p(x) = x²+ x + k
If x – 1 is a factor of p(x), then p(1) = 0
By Factor Theorem
⇒ (1)²+ (1) + k = 0
⇒ 1 + 1 + k = 0
⇒ 2 + k = 0
⇒ k = −2

(ii) p(x) = 2x²+ kx + √2
If x – 1 is a factor of p(x), then p(1)=0
⇒ 2(1)²+ k(1) + √2 = 0
⇒ 2 + k + √2 = 0
⇒ k = −(2 + √2)

(iii) p(x) = kx²– √2x + 1
If x – 1 is a factor of p(x), then p(1)=0
⇒ k(1)²– √2(1) + 1 = 0
⇒ k = √2 – 1

(iv) p(x) = kx²– 3x + k
If x – 1 is a factor of p(x), then p(1) = 0
⇒ k(1)²– 3(1) + k = 0
⇒ k − 3 + k = 0
⇒ 2k − 3 = 0
⇒ k= 3/2

4. Factorize:
(i) 12x²– 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4

Answer –
(i) 12x²– 7x + 1
⇒ 12x² – 4x – 3x + 1
⇒ 4x (3x – 1 ) -1 (3x – 1)
⇒ (3x -1) (4x -1)

(ii) 2x²+7x+3
⇒ 2x²+ 6x + 1x + 3
⇒ 2x (x + 3) + 1(x + 3)
⇒ (2x + 1)(x + 3)

(iii) 6x²+ 5x – 6
⇒ 6x²+ 9x – 4x – 6
⇒ 3x(2x + 3) – 2(2x + 3)
⇒ (2x + 3)(3x – 2)

(iv) 3x²– x – 4
⇒ 3x²– 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)

5. Factorize:
(i) x³– 2x²– x + 2
(ii) x³ – 3x² – 9x – 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² – 2y – 1

Answer –
(i) x³– 2x²– x + 2
Rearranging the terms, we have x³ – x – 2x² + 2
⇒ x(x² – 1) – 2(x² -1)
⇒ (x² – 1)(x – 2)
⇒ [(x)² – (1)²](x – 2) [∵ (a² – b²) = (a + b)(a-b)]
⇒ (x – 1)(x + 1)(x – 2)

(ii) 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 – 5x + x – 5)
⇒ (x + 1)[x(x – 5) + 1(x – 5)]
⇒ (x + 1)(x – 5)(x + 1)

(iii) 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
⇒ 2y³ – 2y² + 3y² – 3y + y – 1
⇒ 2y²(y – 1) + 3y(y – 1) + 1(y – 1)
⇒ (y – 1)(2y² + 3y + 1)
⇒ (y – 1)(2y² + 2y + y + 1)
⇒ (y – 1)[2y(y + 1) + 1(y + 1)]
⇒ (y – 1)(y + 1)(2y + 1)

NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.4