NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4.

Ex 2.4 Class 9 Maths Question 1.
Determine which of the following polynomials has (x + 1) as 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:
The zero of x + 1 is -1.
(i) Let p(x) = x³ + x² + x + 1
∴ p(-1) = (-1)³ + (-1)² + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p(-1) = 0
So, by factor theorem, (x + 1) is a factor of x³ + x² + x + 1.

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

(iii) 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
⇒ p(-1) ≠ 0
So, by factor theorem, (x + 1) is not a factor of x⁴ + 3x³ + 3x² + x + 1.

(iv) 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
⇒ p(-1) ≠ 0
So, by factor theorem, (x + 1) is not a factor of x³ – x² – (2 + √2)x + √2.

Ex 2.4 Class 9 Maths 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) p(x) = x³ – 4x² + x + 6, g(x) = x – 3

Solution:
(i) 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, by factor theorem, g(x) is a factor of p(x).

(ii) 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, by factor theorem, g(x) is not a factor of p(x).

(iii) We have, p(x) = 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 = 0
⇒ p(3) = 0, so, by factor theorem, g(x) is a factor of p(x).

Ex 2.4 Class 9 Maths 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 – √2 x + 1
(iv) p(x) = kx² – 3x + k

Solution:
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

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

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

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

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

Ex 2.4 Class 9 Maths Question 4.
Factorise
(i) 12x² – 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4

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

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

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

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

Ex 2.4 Class 9 Maths Question 5.
Factorise
(i) x³ – 2x² – x + 2
(ii) x³ – 3x² – 9x – 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² – 2y – 1

Solution:
(i) 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) 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) 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) 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)

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