NCERT Solutions for Class 9 Maths Polynomials: Exercise 4

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

(i) x³+x²+x+1

Solution:

p(x) = x³ + x² + x + 1 

Let x+1 be the factor of p(x)
Then x = -1 will be the zero of p(x)
value of p(-1) should be 0
Checking,
=> p(-1) = (-1)³ + (-1)² + (-1) + 1
=> -1 + 1 -1 + 1
=> 0
As p(-1)=0 so (x + 1) is a factor of p(x). 

(ii) x⁴+x³+x²+x+1

Solution:

p(x) = x⁴+x³+x²+x+1
Let x+1 be the factor of p(x)
Then x = -1 will be the zero of p(x)
value of p(-1) should be 0
Checking,
=> p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
=> – 1 + 1 – 1 + 1 -1
=> -1
=> -1 ≠ 0
As p(-1) ≠ 0 so (x + 1) is not a factor of p(x).

(iii) x⁴+3x³+3x²+x+1  

Solution:

p(x) = x⁴+3x³+3x²+x+1  
Let x+1 be the factor of p(x)
Then x = -1 will be the zero of p(x)
value of p(-1) should be 0
Checking,
=> p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
=> 1 – 3 + 3 – 1 + 1
=> -1
=> -1 ≠ 0
As p(-1) ≠ 0 so (x + 1) is not a factor of p(x).

(iv) x³ – x²– (2+√2)x +√2

Solution:

p(x) = x³ – x²– (2+√2)x +√2
Let x+1 be the factor of p(x)
Then x = -1 will be the zero of p(x)
value of p(-1) should be 0
Checking,
=> p(-1) = (-1)³ – (-1)²– (2+√2)(-1) +√2
=> -1 – 1 + 2 + √2 + √2
=> 2√2
=> 2√2 ≠ 0
As p(-1) ≠ 0 so (x + 1) is not a factor of p(x).

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:

p(x) = 2x³+x²– 2x–1
g(x) = x + 1 
By Factor Theorem we know that if x + 1 is a factor of p(x)
Then value of p(-1) should be 0
Checking,
=> p(-1) = 2(-1)³ + (-1)² – 2(-1) -1
=> -2 + 1 + 2 – 1
=> 0
As p(-1) = 0 therefore (x + 1) is a factor of 2x³ + x² – 2x – 1 

(ii) p(x) = x³+3x²+3x+1, g(x) = x+2

Solution:

p(x) = x³+3x²+3x+1 
g(x) = x + 2
By Factor Theorem we know that if x + 2 is a factor of p(x)
Then value of p(-2) should be 0
Checking,
=> p(-2) = (-2)³ + 3(-2)² + 3(-2) +1
=> -8 + 12 – 6 + 1
=> -1
=> -1 ≠ 0
As p(-2) ≠ 0 therefore (x + 2) is not a factor of x³ + 3x² +3x + 1

(iii) p(x)=x³– 4x²+x+6, g(x) = x – 3

Solution:

p(x) = x³– 4x²+x+6 
g(x) = x – 3
By Factor Theorem we know that if x – 3 is a factor of p(x)
Then value of p(3) should be 0
Checking,
=> p(3) = (3)³ – 4(3)² + 3 + 6
=> 27 – 36 + 3 + 6
=> 0
As p(3)=0 so (x – 3) 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:

p(x) = x² + x + k 
By Factor Theorem,
As x-1 is a factor of p(x)
then x = 1 is the zero of p(x)
Therefore p(1) = 0
=> p(1) = (1)² + 1 + k
=> 1 + 1 + k = 0
=> 2 + k = 0
=> k = -2

(ii) p(x) = 2x²+kx+√2

Solution:

p(x) = 2x² + kx + √2
By Factor Theorem,
As x-1 is a factor of p(x)
then x = 1 is the zero of p(x)
Therefore p(1) = 0
=> p(1) = 2(1)² + k(1) + √2
=> 2 + k + √2 = 0
=> 2 + √2 + k = 0
=> k = – (2 + √2) 

(iii) p(x) = kx²–√2x+1

Solution:

p(x) = kx² – √2x + 1
By Factor Theorem,
As x-1 is a factor of p(x)
then x = 1 is the zero of p(x)
Therefore p(1) = 0
=> p(1) = k(1)² – √2(1) + 1
=> k – √2 + 1 = 0
=> k = √2 – 1

(iv) p(x) = kx²–3x+k

Solution:

p(x) = kx² -3x + k
By Factor Theorem,
As x-1 is a factor of p(x)
Then x = 1 is the zero of p(x)
Therefore, p(1) = 0
=> p(1) = k(1)² – 3(1) + k
=> k – 3 + k = 0
=> 2k – 3 = 0
=> k = 3/2

Question 4. Factorize:

(i) 12x²–7x+1

Solution:

p(x) = 12x² – 7x + 1
Using splitting the middle term method,
We need to find a pair of numbers whose sum is -7x
and product is 12x²
-7x can be written as the sum of -3x and -4x
12x² can be written as the product of -3x and -4x
=> 12x² – 7x + 1
=> 12x² -3x -4x +1
=> 3x(4x -1) -1(4x -1)
=> (3x – 1)(4x – 1) are the factors of 12x² – 7x + 1

(ii) 2x²+7x+3

Solution:

p(x) = 2x² + 7x + 3
Using splitting the middle term method,
We need to find a pair of numbers whose sum is 7x
and product is 6x²
7x can be written as the sum of 1x and 6x
6x² can be written as the product of 1x and 6x
=> 2x² + 7x + 3
=> 2x² + 1x + 6x + 3
=> 2x(x + 3) + 1(x + 3)
=> (2x + 1)(x + 3) are the factors of 2x² + 7x + 3

(iii) 6x²+5x-6

Solution:

p(x) = 6x² + 5x – 6
Using splitting the middle term method,
We need to find a pair of numbers whose sum is 5x
and product is -36x²
5x can be written as the sum of 9x and -4x
-36x² can be written as the product of 9x and -4x
=> 6x² + 5x – 6
=> 6x² + 9x – 4x – 6
=> 3x(2x + 3) – 2(2x + 3)
=> (3x – 2)(2x + 3) are the factors of 6x² + 5x – 6

(iv) 3x²–x–4

Solution:

p(x) = 3x² – x – 4
Using splitting the middle term method,
We need to find a pair of numbers whose sum is -x
and product is -12x²
-x can be written as the sum of -4x and 3x
-12x² can be written as the product of -4x and 3x
=> 3x² – x – 4
=> 3x² – 4x + 3x – 4
=> 3x(x + 1) – 4(x + 1)
=> (3x – 4)(x + 1) are the factors of 3x² – x – 4

Question 5. Factorize:

(i) x³–2x²–x+2

Solution:

p(x) = x³– 2x²– x + 2
Factors of 2 are ±1 and ± 2 
Using Hit and Trial Method
p(1) = (1)³ – 2(1)² – (1) + 2
p(1) = 1 – 2 – 1 + 2
p(1) = 0
Therefore, (x – 1) is a factor of x³ – 2x² – x + 2
Performing Long Division :

Dividend = Divisor × Quotient + Remainder
=>p(x) = (x – 1)(x² – x – 2)
=> solving (x² – x -2) 
=> using Splitting the middle term method
=> x² – 2x + x – 2
=> x(x – 2) + 1(x – 2)
=> (x + 1)(x – 2)
=>(x – 1)(x + 1)(x – 2) are the factors of p(x)

(ii) x³–3x²–9x–5

Solution:

p(x) = x³–3x²–9x–5
Factors of -5 are ±1 and ± 5
Using Hit and Trial Method
let x = 1
p(1) = (1)³ – 3(1)² – 9(1) – 5
p(1) = 1 – 3 – 9 -5
p(1) = -16
p(1) ≠ 0
let x = -1
p(-1) = (-1)³ – 3(-1)² – 9(-1) – 5
p(-1) = -1 – 3 + 9 – 5
p(-1) = -9 + 9
p(-1) = 0
Therefore, (x + 1) is a factor of p(x)
Performing Long Division :

Dividend = Divisor × Quotient + Remainder
=>p(x) = (x + 1)(x² – 4x – 5)
=> solving (x² – 4x – 5)
=> using splitting the middle term method
=> x² -5x + x – 5
=> x(x – 5) + 1(x – 5)
=> (x + 1)(x – 5)
=>(x + 1)(x + 1)(x – 5) are the factors of p(x)

(iii) x³+13x²+32x+20

Solution:

p(x) = x³+13x²+32x+20
=> Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
Using Hit and Trial Method
let x = 1
p(1) = (1)³ + 13(1)² + 32(1) + 20
p(1) = 1 + 13 + 32 + 20
p(1) = 66
p(1) ≠ 0
let x = -1
p(-1) = (-1)³ + 13(-1)² + 32(-1) + 20
p(-1) = -1 + 13 – 32 + 20
p(-1) = -33 + 33
p(-1) = 0
Therefore, (x + 1) is a factor of p(x)
Performing Long Division :

Dividend = Divisor × Quotient + Remainder
=> p(x) = (x + 1)(x² + 12x + 20)
=> solving x² + 12x + 20
=> using Splitting the middle term method
=> x² + 10x + 2x + 20
=> x(x + 10) + 2(x + 10)
=> (x + 2)(x + 10) are the factors of x² + 12x + 20
=> (x + 1)(x + 2)(x + 10) are the factors of p(x) 

(iv) 2y³+y²–2y–1

Solution:

p(y) = 2y³+y²–2y–1
=> Factors of -1 are ±1
Using Hit and Trial Method
let x = 1
p(1) = 2(1)³ + (1)² – 2(1) – 1
p(1) = 2 + 1 -2 – 1
p(1) = 0
Therefore, (y – 1) is a factor of p(y)
Performing Long Division :

Dividend = Divisor × Quotient + Remainder
=> p(y) = (y – 1)(2y² + 3y + 1) 
=> solving 2y² + 3y +1
=> using splitting the middle term method
=> 2y² + 2y +y +1
=> 2y(y + 1)+1(2y + 1)
=>(2y + 1)(2y + 1) are the factors of 2y² + 3y + 1
=> (y – 1)(2y + 1)(2y + 1) are the factors of p(y)

NCERT Solutions Class 9 Maths Chapter 2 Polynomials is curated by a team of professionals at GFG to help students in their academic journey. In this article, we have provided the answers to all the questions in the NCERT textbook. All of the problems in this chapter’s exercise from the NCERT textbook for Class 9 are covered in the NCERT Solutions for Class 9 Maths.

The topics covered in the Polynomials chapter in Class 9 are Polynomials, Types of polynomials, Operations on polynomials, Degree of a polynomial, Factorization of polynomials, Factor Theorem, Remainder theorem, Algebraic identities and Application problems are the important Chapters in Class 9 Maths.