NCERT Solutions Class 9th Maths Chapter – 2 Polynomials Exercise – 2.3

Question 1. Determine which of the following polynomials has (x + 1) a factor:

(i) x³ + x² + x + 1

Solution
The zero of x + 1 is -1
Let p(x) = x³ + x² + x + 1
p(-1) = (-1)³ + (-1)² + (-1) + 1
= -1 + 1 – 1 + 1
⇒ 0
So, (x+ 1) is a factor of x³ + x² + x + 1

(ii) x⁴ + x³ + x² + x + 1

Solution
Let p(x) = x⁴ + x³ + x² + x + 1
P(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
= 1 – 1 + 1 – 1 + 1
⇒ 1 ≠ 0
So, (x + 1) is not a factor of x⁴ + x³ + x² + x+ 1

(iii) x⁴ + 3x³ + 3x² + x + 1

Solution
Let p(x) = x⁴ + 3x³ + 3x² + x + 1
p(-1)= (-1)⁴ + 3 (-1)³ + 3 (-1)² + (- 1) + 1
= 1 – 3 + 3 – 1 + 1
⇒ 1 ≠ 0
So, (x + 1) is not a factor of x⁴ + 3x³ + 3x² + x+ 1

(iv) x³ – x² – (2 +√2 )x + √2

Solution
Let p(x) = x³ – x² – (2 + √2) x + √2
p(- 1) = (- 1)3- (-1)2 – (2 + √2)(-1) + √2
= -1 – 1 + 2 + √2 + √2
= 2√2 ≠ 0
So, (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

Solution
We have, p(x)= 2x³ + x² – 2x – 1 and g(x) = x + 1
∴ p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= 2(-1) + 1 + 2 – 1
= -2 + 1 + 2 -1
= 0
⇒ p(-1) = 0, so g(x) is a factor of p(x)

(ii) p(x)= x³ + 3x² + 3x + 1, g(x) = x + 2

Solution
We have, p(x) x³ + 3x² + 3x + 1 and g(x) = x + 2
p(-2) = (-2)³ + 3(-2)²+ 3(-2) + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x)

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

Solution
We have, = x³ – 4x² + x + 6 and g (x) = x – 3
p(3) = (3)³ – 4(3)² + 3 + 6
= 27 – 4(9) + 3 + 6
= 27 – 36 + 3 + 6
= -36 + 36
= 0
⇒ p(3) = 0, so 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:

(i) p(x) = x² + x + k

Solution
Here, p(x) = x² + x + k
Since, p(1) = (1)² + 1 + k
⇒ 1 + 1 + k = 0
⇒ k + 2 = 0
⇒ k = -2

(ii) p(x) = 2x² + kx + √2

Solution
Here, p(x) = 2x² + kx + √2
Since, p(1) = 2(1)² + k(1) + √2
= 2 + k + √2 = 0
⇒ k = -2 – √2
⇒ k = -(2 + √2)

(iii) p(x) = kx² – √2 x + 1

Solution
Here, p(x) = kx² – √2 x + 1
Since, p(1) = k(1)² – (1) + 1
⇒ k – √2 + 1 = 0
⇒ k = √2 -1

(iv) p(x) = kx² – 3x + k

Solution
Here, p(x) = kx² – 3x + k
p(1) = k(1)² – 3(1) + k
⇒ k – 3 + k
⇒ 2k – 3 = 0
⇒ k = 3/4

Question 4. Factorise:

(i) 12x² – 7x + 1

Solution
We have, 12x² – 7x + 1
= 12x² – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)

(ii) 2x² + 7x + 3

Solution
We have, 2x² + 7x + 3
= 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1) (x + 3)

(iii) 6x² + 5x – 6

Solution
We have, 6x² + 5x – 6
= 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3) (3x – 2)

(iv) 3x² – x – 4

Solution
We have, 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4) (x + 1)

Question 5. Factories:

(i) x³ – 2x² – x + 2

Solution
We have, 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)
= (x – 1)(x + 1)(x – 2)
[∵ (a² – b²) = (a + b)(a-b)]
Thus, x³ – 2x² – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) x³ – 3x² – 9x – 5

Solution
We have, 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)
Thus, x³ – 3x² – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) x³ + 13x² + 32x + 20

Solution
We have, 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)
Thus, x³ + 13x² + 32x + 20
= (x + 1)(x + 2)(x + 10)

(iv) 2y³ + y² – 2y – 1

Solution
We have, 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)
Thus, 2y³ + y² – 2y – 1
= (y – 1)(y + 1)(2y +1)