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 part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4.

• Polynomials Class 9 Ex 2.1
• Polynomials Class 9 Ex 2.2
• Polynomials Class 9 Ex 2.3
• Polynomials Class 9 Ex 2.4
• Polynomials Class 9 Ex 2.5

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 2
Chapter Name	Polynomials
Exercise	Ex 2.4
Number of Questions Solved	5
Category	NCERT Solutions

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) a factor:
(i) x³ +x² + 1
(ii) x⁴ + x³+x²+x+1
(iii) x⁴ + 3x³ + 3x² + x + 1
(iv) x³ -x² -(2 + – )x + 
Solution:
The zero of x +1 is -1.
(i) Let p(x) = x³ + x² + x +1
Then,   p(-1) =  (-1)³ + (-1)² + (-1) +1
= -1+1-1+1 => p(-1)=   0
So, by the Factor theorem (x +1) is a factor of x³ + x² + x+1.

(ii) Let p(x)= x⁴ + x³ + x² + x +1
Then,   p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) +1
=1-1+1-1+1 =>      p(-1) =  1
So, by the.Factor theorem (x +1) is not a factor of x⁴ + x³ + x² + x +1

(iii) Let p(x) = x⁴ + 3x³ + 3x² + x +1
Then,  p(-l) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) +1
=1-3+3-1+1
=>  P(-1) =  1
So, by the Factor theorem (x +1) is not a factor of x⁴ + 3x³ + 3x² + x +1

(iv) Let p(x) = x³-x²-(2 +  )x + 
Then, p(-1) = (-1)³ – (-1)² – (2 +  )(-1) + 
= —1—1+2 +  + 
= 2
So, by the Factor theorem (x + 1) is not a factor ofx³ -x² -(2 +  x + 

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) The zero of g(x) = x +1 is x = -1.
Then,  p(-1)= 2(-1)³ +(-1)² -2(-1)-1
[∵  p(x) = 2x³ + x² -2x-1]
= -2 +1+2-1 ⇒ p(-1) = 0
Hence, g(x) is a factor of p(x).

(ii) The zero of g(x) = x + 2 is -2.
Then, p(-2) = (-2)³ + 3(-2)² + 3(-2) +1
[∵ p(x) = x³ + 3x² + 3x +1]
= -8+12-6 +1 =>p(-2)=-1
Hence, g(x) is not a factor of p(x).

(iii) The zero of g(x) = x – 3 is 3.
Then,p(3) = 3³ – 4(3)² +3 + 6
[∵ p(x) = x³ – 4x² + x + 6]
= 27-36 +3 + 6 =»       p(3) = 0
Hence, 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 + V2=
(iii) p(x) = kx² -42x + 1
(iv) p(x) = kx² -3x + k
Solution:

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:

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) Let p(x) = x³-2x²-x+2, constant term of P(x) is 2.
Factors of 2 are ± 1 and ± 2.
Now,  p(1)=1³-2(1)²-1+2
=1-2-1+2
By trial we find that p(l) = 0, so (x -1) is a factor of p(x).
So, x³-2x²-x+2= x³-x²-x² + x-2x+2
= x²(x-1)-x(x-1)-2(x-1)
= (x—1)(x² -x-2)
= (x -1)(x² -2x + x -2)
= (x -1) [x(x -2) + l(x -2)]
= (x-1)(x-2)(x+1)

(ii) Let  p(x)= x³ -3x² -9x-5
By trial, we find that p(5) = (5)³ – 3(5)² – 9(5) – 5
= 125 – 75 – 45 – 5 = 0
So, (x – 5) is a factor of p(x).
So, x³-3x²-9x-5=x³-5x²+2x² -10x + x – 5
= x² (x – 5) + 2x(x – 5) + 1(x – 5)
= (x – 5) (x²+2x+1)
= (x – 5) (x² + x + x +1)
= (x + 5) [x(x +1) + 1(x +1)]
= (x-5)(x +1) (x +1)
= (x – 5)(x +1)²

(iii) Let p(x)= x³ +13x² +32x +20
By trial, we find that p(-1) = (-1)³ +13(-1)² + 32(-1) + 20
= -1 +13 -32 + 20 = -33 + 33 = 0
So (x +1) is a factor of p(x).
So, 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 + 10x + 2x + 20)
= (x +1)[x(x +10) + 2(x +10)]
= (x +l)(x +10)(x + 2)

(iv) Let p(y) = 2y³ + y² – 2y -1
By trial we find that p(l) = 2(1)³ + (1)² -2(1) -1,
=2+1-2-1=0 So (y -1) is a factor of p(y).
So, 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) + l(y +1)
= (y-1)(y+ l)(2y+ 1)

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