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.
| 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
Question 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
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 (-1)= (-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 + √2) x + √2
Then, p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2
= -1 – 1+ 2 +√2+√2
= 2√2
So, by the Factor theorem (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
(ii) p(x)= x3 + 3x2 + 3x + X 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).
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:
The zero of x – 1 is 1.
(i) (x – 1) is a factor of p (x),then p(1)= 0 (By Factor theorem)
⇒ 12 + 1 + k = 0 [∵ p(x) = x2 + x + k]
⇒ 2 + k =0
⇒ k = -2
(ii) ∵ (x -1) is a factor of p (x), then p (1) = 0 (By Factor theorem)
⇒ 2(1)2 + k(1)+√2= 0 [∵p(x) = 2x2 + kx+ -√2]
⇒ 2 + k + √2 = 0
⇒ k = – (2 + √2)
(iii) ∵ (x-1) is a factor of p (x), then p (1) = 0 (By Factor theorem)
⇒ k (1)2 – √2 + 1 = 0 [∵p(x) = kx2 – √2x + 1]
⇒ k = (√2 – 1)
(iv) ∵ (x-1) is a factor of p (x), then p (1) = 0 (By Factor theorem)
⇒ k(1)2 – 3 + k = 0 [∵p(x) = kx2 – 3x + k]
⇒ 2k-3 = 0
⇒ k = \(\frac { 3 }{ 2 }\)
Question 4.
Factorise
(i) 12x2 – 7x +1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4
Solution:
(i) 12x2 – 7x + 1 = 12x2 – 4x- 3x + 1 (Splitting middle term)
= 4x (3x – -0 -1 (3x-1)
= (3x -1) (4x -1)
(ii)2x2 + 7x + 3 = 2x2 + 6x + x + 3 (Splitting middle term)
= 2x (x + 3) +1 (x + 3) = (x + 3) (2x+ 1)
(iii) 6x2 + 5x – 6= 6x2 + 9x- 4x- 6 (Splitting middle term)
= 3x(2x+3)-2(2x+3)=(2x+3)(3x-2)
(iv) 3x2 – x- 4= 3x2-4x+3x-4 (Splitting middle term)
= x (3x – 4) + 1 (3x – 4)= (3x- 4) (x + 1)
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
p(1) = 0
By trial, we find that p (1) = 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) + 1 (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+ 1) (x+ 10) (x + 2)
(iv) Let p (y) = 2y3 + y2 – 2y -1
By trial we find that p(1) = 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) + 1 (y + 1)]
= (y – 1)(y+1)(2y+1)
We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4, drop a comment below and we will get back to you at the earliest.