Category: CBSE Class 9

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.

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.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) x – \(\frac { 1 }{ 2 }\)
(iii) x
(iv) x + π
(v) 5 + 2x
Solution:
Let p (x) = x³ + 3x² + 3x + 1
(i) The zero of x+ 1 is – 1. (∵ x+ 1 = 0= x = -1)

Question 2.
Find the remainder when x³-ax² +6x-ais divided by x – a.
Solution:
The zero of x – a is a. (∵ x – a = 0 =» x= a)
Let p (x) = x³ – ax² + 6x – a
So, p (a) = a³ – a (a)² + 6a – a = a³ – a³ + 5a
⇒ p (a) = 5a
∴ Required remainder = 5a (By Remainder theorem)

Question 3.
Check whether 7 + 3x is a factor of 3x³+7x.
Solution:
Let f(x) = 3x³+7x

We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the value of the polynomial 5x -4x² + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
Let p (x) = 5x – 4x²+ 3
(i) The value of p (x) = 5x – 4x²+ 3 at x= 0 is
p(0) = 5 x 0 – 4 x 0²+3
⇒ P (0) = 3
(ii) The value of p (x) = 5x – 4x² + 3 at x = -1 is
p(-1) = 5(-D-4(-1)² + 3 = – 5 -4 + 3
⇒ P(-1) = -6
(iii) The value of p (x) = 5x- 4x² + 3 at x = 2 is
p (2) = 5 (2)- 4 (2)² + 3= 10- 16+ 3
⇒ P (2) = – 3

Question 2.
Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y² – y +1
(ii) p (t) = 2 +1 + 2t² -t³
(iii) P (x) = x³
(iv) p (x) = (x-1) (x+1)
Solution:
(i) p (y) = y² -y +1
∴ p(0) = 0²-0+1
⇒ p(0) = 1
p(1) = 1²-1+ 1
⇒ p(1) = 1
and p (2) = 2² – 2 + 1 =4-2+1
⇒ P (2) = 3
(ii) p (t) = 2 + t + 2t² -1³
p(0) = 2+ 0+ 2 x 0²– 0³
⇒ P (0) = 2
p (1) = 2 + 1 + 2 x 1² – 1³
⇒ p (1) = 3 + 2 – 1
⇒ p(1) = 4
and p (2) = 2 + 2 + 2 x 2² – 2³
=4+8-8
⇒ P (2) = 4
(iii) P(x) = x³
⇒ p (0) = 0³ ⇒ p (0) = 0 ⇒ p (1) = 1³
⇒ P (1) = 1
and p (2) = 2³ ⇒ p (2) = 8
(iv) p(x) = (x-1)(x+ 1)
p(0) = (0-1)(0+1)
⇒ P (0) = – 1
p (1) = (1 – 1) (1 + 1)
⇒ P (1) = 0
and p (2) = (2-1) (2+1)
⇒ P (2) = 3

Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them.
(i)p(x) = 3x + 1,x = –\(\frac { 1 }{ 3 }\)
(ii)p (x) = 5x – π, x = \(\frac { 4 }{ 5 }\)
(iii) p (x) = x² – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x², x = 0
(vi) p (x) = lx + m, x = – \(\frac { m }{ l }\)
(vii) P (x) = 3x² – 1, x = – \(\frac { 1 }{ \sqrt { 3 } }\),\(\frac { 2 }{ \sqrt { 3 } }\)
(viii) p (x) = 2x + 1, x = \(\frac { 1 }{ 2 }\)
Solution:

Question 4.
Find the zero of the polynomial in each of the following cases
(i) p(x)=x+5
(ii) p (x) = x – 5
(iii) p (x) = 2x + 5
(iv) p (x) = 3x – 2
(v) p (x) = 3x
(vi) p (x)= ax, a≠0
(vii) p (x) = cx + d, c≠ 0 where c and d are real numbers.
Solution:
(i) We have, p (x)= x+ 5
Now, p (x) = 0
⇒ x+ 5 = 0
⇒ x = -5
∴ – 5 is a zero of the polynomial p (x).
(ii) We have, p (x) = x – 5
Now, p (x) = 0
⇒ x – 5 = 0
⇒ x = 5
∴ 5 is a zero of the polynomial p (x).
(iii) We have, p (x) = 2x + 5
Now, P (x)= 0
⇒ 2x+ 5= 0
⇒ x = –\(\frac { 5 }{ 2 }\)
∴ –\(\frac { 5 }{ 2 }\) is a zero of the polynomial p (x).
(iv) We have, p (x)= 3x- 2
Now p(x) = 0
⇒ 3x- 2 = 0
⇒ x= \(\frac { 2 }{ 3 }\)
∴ \(\frac { 2 }{ 3 }\) is a zero of the polynomial p (x).
(v) We have, p (x) = 3x
Now, p (x)= 0
⇒ 3x=0
⇒ x =0
∴ 0 is a zero of the polynomial p (x).
(vi) We have, p (x)= ax, a ≠ 0
Now, p (x)= 0 ⇒ ax= 0
⇒ x= 0
∴ 0 is.a zero of the polynomial p (x).
(vii) We have, p (x) = cx + d,c ≠ 0
Now, p (x) = 0
⇒ cx + d = 0
x = – \(\frac { d }{ c }\)
∴ – \(\frac { d }{ c }\) is a zero of the polynomial p (x).

We hope the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2, drop a comment below and we will get back to you at the earliest.