NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

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.

These important NCERT Solutions for Class 9 Maths hold a significant weightage of 12 marks in the Class 9 Maths CBSE examination. It covers essential topics including:

• Polynomials in One Variable
• Zeroes of a Polynomial
• Remainder Theorem
• Factorisation of Polynomials
• Algebraic Identities

To excel in this chapter, students can utilize NCERT Solutions for Class 9. These resources are invaluable for mastering concepts and preparing effectively for their Class 9 Maths exams.

Question 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) 4x² – 3x + 7
(ii) y² + √2
(iii) 3√t + t√2
(iv) y + 2/y
(v) x¹⁰ + y³ + t⁵⁰

Solution: 

(i) The algebraic expression 4x² – 3x + 7 can be written as 4x² – 3x + 7x⁰

As we can see, all exponents of x are whole numbers, 

So, the given expression 4x² – 3x + 7 is polynomial in one variable.

(ii) The algebraic expression y² + √2 can be written as y² + √2y⁰

As we can see, all exponents of y are whole numbers, 

So, the given expression y² + √2 is polynomial in one variable.

(iii) The algebraic expression 3 √t + t√2 can be written as 3 t^1/2 + √2.t

As we can see, one exponent of t is 1/2, which is not a whole number,

So, the given expression 3 √t + t√2 is not a polynomial in one variable.

(iv) The algebraic expression y + 2/y can be written as y + 2.y^-1

As we can see, one exponent of y is -1, which is not a whole number,

So, the given expression y+ 2/y is not a polynomial in one variable.

(v) The given algebraic expression is x¹⁰+ y³+ t⁵⁰

As we can see, the expression contains three variables i.e x, y, and t,

So, the given expression x¹⁰ + y³ + t⁵⁰ is not a polynomial in one variable.

Question 2. Write the coefficients of x² in each of the following

(i) 2 + x² + x
(ii) 2 – x² + x³
(iii) pi/2 x² + x
(iv) √2x – 1

Solution: 

(i) The given algebraic expression is 2 + x² + x

As we can clearly see, the coefficient of x² is 1.

(ii) The given algebraic expression is 2 – x² + x³

As we can clearly see, the coefficient of x² is -1.

(iii) The given algebraic expression is pi/2 x² + x

As we can clearly see, the coefficient of x² is pi/2.

(iv) The given algebraic expression is √2 x — 1

As we can clearly see, the coefficient of x² is 0.

Question 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution: 

A Binomial having degree 35 is 4x³⁵ + 50

A Monomial having degree 100 is 3t¹⁰⁰pi

Question 4. Write the degree of each of the following polynomials

(i) 5x³ + 4x² + 7x
(ii) 4 – y²
(iii) 5t – √7
(iv) 3

Solution:  

The highest power of a variable in the given expression is known as the Degree of the polynomial 

(i) The given expression is 5x³ + 4x² + 7x

As we can clearly see, the highest power of variable x is 3,

So, the degree of given polynomial 5x³+4x² + 7x is 3.

(ii) The given expression is 4 – y²

As we can clearly see, the highest power of variable y is 2,

So, the degree of given polynomial 4 – y² is 2.

(iii) The given expression is 5t – √7

As we can clearly see, the highest power of variable t is 1,

So, the degree of given polynomial 5t – √7 is 1.

(iv) The given expression 3 can be written as 3x⁰

As we can clearly see, the highest power of variable x is 0,

So, the degree of given polynomial 3 is 0.

Question 5. Classify the following as linear, quadratic, and cubic polynomials

(i) x² + x
(ii) x – x³
(iii) y + y² + 4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x³

Solution: 

(i) Since the degree of given polynomial x² + x is 2,

So, it is a Quadratic Polynomial.

(ii) Since the degree of given polynomial x – x³ is 3,

So, it is a Cubic Polynomial.

(iii) Since the degree of given polynomial y + y² + 4 is 2,

So, it is a Quadratic Polynomial.

(iv) Since the degree of given polynomial 1 + x is 1,

So, it is a Linear Polynomial.

(v) Since the degree of given polynomial 3t is 1, 

So, it is a Linear Polynomial.

(vi) Since the degree of given polynomial r² is 2,

So, it is a Quadratic Polynomial.

(vii) Since the degree of given polynomial 7x³ is 3,

So, it is a Cubic Polynomial.

Question 1: Find the value of the polynomial (x) = 5x − 4x² + 3  

(i) x = 0

(ii) x = –1

(iii) x = 2

Solution:

Given equation: 5x − 4x² + 3

Therefore, let f(x) = 5x – 4x² + 3

(i) When x = 0

f(0) = 5(0)-4(0)²+3

= 3

(ii) When x = -1

f(x) = 5x−4x²+3

f(−1) = 5(−1)−4(−1)²+3

= −5–4+3

= −6

(iii) When x = 2

f(x) = 5x−4x²+3

f(2) = 5(2)−4(2)²+3

= 10–16+3

= −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+t+2t²−t³

(iii) p(x) = x³

(iv) P(x) = (x−1)(x+1)

Solution:

(i) p(y) = y² – y + 1

Given equation: p(y) = y²–y+1

Therefore, p(0) = (0)²−(0)+1 = 1

p(1) = (1)²–(1)+1 = 1

p(2) = (2)²–(2)+1 = 3

Hence, p(0) = 1, p(1) = 1, p(2) = 3, for the equation p(y) = y²–y+1

(ii) p(t) = 2 + t + 2t² − t³

Given equation: p(t) = 2+t+2t²−t³

Therefore, p(0) = 2+0+2(0)²–(0)³ = 2

p(1) = 2+1+2(1)²–(1)³ = 2+1+2–1 = 4

p(2) = 2+2+2(2)²–(2)³ = 2+2+8–8 = 4

Hence, p(0) = 2,p(1) =4 , p(2) = 4, for the equation p(t)=2+t+2t²−t³

(iii) p(x) = x³

Given equation: p(x) = x³

Therefore, p(0) = (0)³ = 0

p(1) = (1)³ = 1

p(2) = (2)³ = 8

Hence, p(0) = 0,p(1) = 1, p(2) = 8, for the equation p(x)=x³

(iv) p(x) = (x−1)(x+1)

Given equation: p(x) = (x–1)(x+1)

Therefore, p(0) = (0–1)(0+1) = (−1)(1) = –1

p(1) = (1–1)(1+1) = 0(2) = 0

p(2) = (2–1)(2+1) = 1(3) = 3

Hence, p(0)= 1, p(1) = 0, p(2) = 3, for the equation p(x) = (x−1)(x+1)

Question 3: Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x+1, x=−1/3

(ii) p(x) = 5x–π, x = 4/5

(iii) p(x) = x²−1, x=1, −1

(iv) p(x) = (x+1)(x–2), x =−1, 2

(v) p(x) = x², x = 0

(vi) p(x) = lx+m, x = −m/l

(vii) p(x) = 3x²−1, x = -1/√3 , 2/√3

(viii) p(x) = 2x+1, x = 1/2

Solution:

(i) p(x)=3x+1, x=−1/3

Given: p(x)=3x+1 and x=−1/3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/3

p(−1/3) = 3(-1/3)+1

= −1+1 

= 0

Hence, p(x) of -1/3 = 0

(ii) p(x)=5x–π, x = 4/5

Given: p(x)=5x–π and x = 4/5

Therefore, substituting the value of x in equation p(x), we get.

For, x = 4/5

p(4/5) = 5(4/5)–π 

= 4–π

Hence, p(x) of 4/5 ≠ 0

(iii) p(x)=x²−1, x=1, −1

Given: p(x)=x²−1 and x=1, −1

Therefore, substituting the value of x in equation p(x), we get.

For x = 1 

p(1) = 1²−1

=1−1 

= 0

For, x = -1

p(−1) = (-1)²−1 

= 1−1 

= 0

Hence, p(x) of 1 and -1 = 0

(iv) p(x) = (x+1)(x–2), x =−1, 2

Given: p(x) = (x+1)(x–2) and x =−1, 2

Therefore, substituting the value of x in equation p(x), we get.

For, x = −1

p(−1) = (−1+1)(−1–2)

= (0)(−3) 

= 0

For, x = 2

p(2) = (2+1)(2–2) 

= (3)(0) 

= 0

Hence, p(x) of −1, 2 = 0

(v) p(x) = x², x = 0

Given:  p(x) = x² and x = 0

Therefore, substituting the value of x in equation p(x), we get.

For, x = 0

p(0) = 0² = 0

Hence, p(x) of 0 = 0

(vi) p(x) = lx+m, x = −m/l

Given: p(x) = lx+m and x = −m/l

Therefore, substituting the value of x in equation p(x), we get.

For, x = −m/l

p(-m/l)= l(-m/l)+m 

= −m+m 

= 0

Hence, p(x) of -m/l = 0

(vii) p(x) = 3x²−1, x = -1/√3 , 2/√3

Given: p(x) = 3x²−1 and x = -1/√3 , 2/√3

Therefore, substituting the value of x in equation p(x), we get.

For, x = -1/√3

p(-1/√3) = 3(-1/√3)² -1 

= 3(1/3)-1 

= 1-1 

= 0

For, x =  2/√3

p(2/√3) = 3(2/√3)² -1 

= 3(4/3)-1 

= 4−1

=3 ≠ 0

Hence, p(x) of -1/√3 = 0

but, p(x) of 2/√3 ≠ 0

(viii) p(x) =2x+1, x = 1/2

Given: p(x) =2x+1 and x = 1/2

Therefore, substituting the value of x in equation p(x), we get.

For, x = 1/2

p(1/2) = 2(1/2)+1 

= 1+1 

= 2≠0

Hence, p(x) of 1/2 ≠ 0

Question 4: Find the zero of the polynomials 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  

(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.

Solution:

(i) p(x) = x+5  

Given: p(x) = x+5

To find the zero, let p(x) = 0

p(x) = x+5

0 = x+5

x = −5

Therefore, the zero of the polynomial p(x) = x+5 is when x = -5

(ii) p(x) = x–5

Given: p(x) = x–5

p(x) = x−5

x−5 = 0

x = 5

Therefore, the zero of the polynomial p(x) = x–5 is when x = 5

(iii) p(x) = 2x+5

Given: p(x) = 2x+5

p(x) = 2x+5

2x+5 = 0

2x = −5

x = -5/2

Therefore, the zero of the polynomial p(x) = 2x+5 is when x = -5/2

(iv) p(x) = 3x–2  

Given: p(x) = 3x–2  

p(x) = 3x–2

3x−2 = 0

3x = 2

x = 2/3

Therefore, the zero of the polynomial p(x) = 3x–2 is when x = 2/3

(v) p(x) = 3x  

Given: p(x) = 3x  

p(x) = 3x

3x = 0

x = 0

Therefore, the zero of the polynomial p(x) = 3x   is when x = 0

(vi) p(x) = ax, a0

Given: p(x) = ax, a≠ 0

p(x) = ax

ax = 0

x = 0

Therefore, the zero of the polynomial p(x) = ax is when x = 0

(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.

Given: p(x) = cx+d

p(x) = cx + d

cx+d =0

x = -d/c

Therefore, the zero of the polynomial p(x) = cx+d is when x = -d/c

Question 1. Find the remainder when x³ + 3x² + 3x + 1 is divided by

(i) x + 1

Solution:

x + 1 = 0

x  = −1

Therefore remainder will be f(x):

f(−1) = (−1)³ + 3(−1)² + 3(−1) + 1

= −1 + 3 − 3 + 1

= 0

(ii) x – 1/2

Solution:

x – 1/2 = 0

x = 1/2

Therefore remainder will be f(x):

f(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1

= (1/8) + (3/4) + (3/2) + 1

= 27/8

(iii) x

Solution:

x = 0

Therefore remainder will be f(x):

f(0) = (0)³ + 3(0)² + 3(0) + 1

= 1

(iv) x + pi

Solution:

x + pi = 0

x = −pi

Therefore remainder will be f(x):

f(−pi) = (−pi)³ + 3(−pi)² + 3(−pi) + 1

= −pi³ + 3pi² − 3pi + 1

(v) 5 + 2x

Solution:

5 + 2x = 0

2x = −5

 x = -5/2

Therefore remainder will be f(x) :

f(-5/2) = (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1 

= (-125/8) + (75/4) – (15/2) + 1

= -27/8

Question 2. Find the remainder when x³ − ax² + 6x − a is divided by x – a.

Solution:

Let f(x) = x³ − ax² + 6x − a

x − a = 0

∴ x = a

Therefore remainder will be f(x):

f(a) = (a)³ − a(a²) + 6(a) − a

= a³ − a³ + 6a − a 

= 5a

Question 3. Check whether 7 + 3x is a factor of 3x³ + 7x.

Solution:

7 + 3x = 0

3x = −7

 x = -7/3

Therefore remainder will be f(x):

f(-7/3) = 3(-7/3)³ + 7(-7/3) 

= – (343/9) + (-49/3)

= (-343- (49) * 3)/9

= (-343 – 147)/9

= – 490/9 ≠ 0

∴ 7 + 3x is not a factor of 3x³ + 7x

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)

Question 1. Use suitable identities to find the following products:

(i) (x + 4) (x + 10) 

Solution:

Using formula, (x + a) (x + b) = x² + (a + b)x + ab

[So, a = 4 and b = 10]

(x + 4) (x + 10) = x² + (4 + 10)x + (4 × 10)

= x² + 14x + 40

(ii) (x + 8) (x – 10)      

Solution:

Using formula, (x + a) (x + b) = x² + (a + b)x + ab

[So, a = 8 and b = −10]

(x + 8) (x – 10) = x² + (8 + (-10) )x + (8 × (-10))

= x² + (8 – 10) x – 80

= x² − 2x – 80

(iii) (3x + 4) (3x – 5)

Solution:

Using formula, (y + a) (y + b) = y² + (a + b)y + ab

[So, y = 3x, a = 4 and b = −5]

(3x + 4) (3x − 5) = (3x)² + [4 + (-5)]3x + 4 × (-5)

= 9x² + 3x (4 – 5) – 20

= 9x² – 3x – 20

(iv) (y² + ) (y² – )

Solution:

Using formula, (a + b) (a – b) = a² – b²

[So, a = y² and b = ]

(y ² + ) (y² – ) = (y²)² – ()^2

= y ⁴ – 

Question 2. Evaluate the following products without multiplying directly:

(i) 103 × 107

Solution:

103 × 107 = (100 + 3) × (100 + 7)

Using formula, (x + a) (x + b) = x² + (a + b)x + ab

Then,

x = 100

a = 3

b = 7

So, 103 × 107 = (100 + 3) × (100 + 7)

= (100)² + (3 + 7)100 + (3 × 7)

= 10000 + 1000 + 21

= 11021

(ii) 95 × 96  

Solution:

95 × 96 = (100 – 5) × (100 – 4)

Using formula, (x – a) (x – b) = x² – (a + b)x + ab

Then, According to the identity

x = 100

a = 5

b = 4

So, 95 × 96 = (100 – 5) × (100 – 4)

= (100)² – 100 (5+4) + (5 × 4)

= 10000 – 900 + 20

= 9120

(iii) 104 × 96

Solution:

104 × 96 = (100 + 4) × (100 – 4)

Using formula, (a + b) (a – b) = a² – b²

Then,

a = 100

b = 4

So, 104 × 96 = (100 + 4) × (100 – 4)

= (100)² – (4)²

= 10000 – 16

= 9984

Question 3. Factorize the following using appropriate identities:

(i) 9x² + 6xy + y²

Solution:

9x² + 6xy + y² = (3x)² + (2 × 3x × y) + y²

Using formula, a² + 2ab + b² = (a + b)²

Then, 

a = 3x

b = y

9x² + 6xy + y² = (3x)² + (2 × 3x × y) + y²

= (3x + y)²

= (3x + y) (3x + y)

(ii) 4y² − 4y + 1

Solution:

4y² − 4y + 1 = (2y)² – (2 × 2y × 1) + 1

Using formula, a² – 2ab + b² = (a – b)²

Then,

a = 2y

b = 1

= (2y – 1)²

= (2y – 1) (2y – 1)

(iii)  x² – 

Solution:

x² –  = x² – 

Using formula, a² – b² = (a – b) (a + b)

Then, 

a = x

b = 

= (x – ) (x + )

Question 4. Expand each of the following, using suitable identities:

(i) (x + 2y + 4z)²

Solution:

Using formula, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Then, 

x = x

y = 2y

z = 4z

(x + 2y + 4z)² = x² + (2y)² + (4z)² + (2 × x × 2y) + (2 × 2y × 4z) + (2 × 4z × x)

= x² + 4y² + 16z² + 4xy + 16yz + 8xz

(ii) (2x − y + z)²  

Solution:

Using formula, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Then, 

x = 2x

y = −y

z = z

(2x − y + z)² = (2x)² + (−y)² + z² + (2 × 2x × −y) + (2 × −y × z) + (2 × z × 2x)

= 4x² + y² + z² – 4xy – 2yz + 4xz

(iii) (−2x + 3y + 2z)²

Solution:

Using formula, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Then, 

x = −2x

y = 3y

z = 2z

(−2x + 3y + 2z)² = (−2x)² + (3y)² + (2z)² + (2 ×−2x × 3y) + (2 ×3y × 2z) + (2 ×2z × −2x)

= 4x² + 9y² + 4z² – 12xy + 12yz– 8xz

(iv) (3a – 7b – c)²

Solution:

Using formula, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Then, 

x = 3a

y = – 7b

z = – c

(3a – 7b –  c)² = (3a)² + (– 7b)² + (– c)² + (2 × 3a × – 7b) + (2 × –7b × –c) + (2 × –c × 3a)

= 9a² + 49b² + c² – 42ab + 14bc – 6ca

(v) (–2x + 5y – 3z)²

Solution:

Using formula, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Then, 

x = –2x

y = 5y

z = – 3z

(–2x + 5y – 3z)² = (–2x)² + (5y)² + (–3z)² + (2 × –2x × 5y) + (2 ×  5y × – 3z) + (2 × –3z × –2x)

= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

(vi) (a – b + 1)²

Solution:

Using formula, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Then, 

x = a

y = b

z = 1

(a – ()b + 1)² = [a]² + [b]² + 1² + [2 x }a x b] + [2 xb x 1] + [2 x 1 x  a]

= a² + b² + 1 – ab – b + a

Question 5. Factorize:

(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

Solution:

Using formula, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Then, x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²

4x² + 9y² + 16z² + 12xy – 24yz – 16xz = (2x)² + (3y)² + (−4z)2 + (2 × 2x × 3y) + (2 × 3y × −4z) + (2 × −4z × 2x)

= (2x + 3y – 4z)²

= (2x + 3y – 4z) (2x + 3y – 4z)

(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz

Solution:

Using formula, (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Then, x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²

2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz

= (-√2x)² + (y)² + (2√2z)² + (2 × -√2x × y) + (2 × y × 2√2z) + (2 × 2√2 × −√2x)

= (−√2x + y + 2√2z)²

= (−√2x + y + 2√2z) (−√2x + y + 2√2z)

Question 6. Write the following cubes in expanded form:

(i) (2x + 1)³

Solution:

Using formula,(x + y)³ = x³ + y³ + 3xy(x + y)

(2x + 1)³= (2x)³ + 1³ + (3 × 2x ×1) (2x + 1)

= 8x³ + 1 + 6x(2x + 1)

= 8x³ + 12x² + 6x + 1

(ii) (2a − 3b)³

Solution:

Using formula, (x – y)³ = x³ – y³ – 3xy(x – y)

(2a − 3b)³ = (2a)³ − (3b)³ – (3 × 2a × 3b) (2a – 3b)

= 8a³ – 27b³ – 18ab(2a – 3b)

= 8a³ – 27b³ – 36a²b + 54ab²

(iii) (x + 1)³

Solution:

Using formula, (x + y)³ = x³ + y³ + 3xy(x + y)

(x+ 1)³ = (x)³ + 1³ + (3 × x × 1) (x + 1)

= x³ + 1 + x² + x

= 

(iv) (x − y)³

Solution:

Using formula, (x – y)³ = x³ – y³ – 3xy(x – y)

(x − y)³ = x³ − [y]³ – 3(x) y[x − y]

= x³ –y³ – 2x²y + xy²

Question 7. Evaluate the following using suitable identities:  

(i) (99)³

Solution:

99 = 100 – 1

Using formula, (x – y)³ = x³ – y³ – 3xy(x – y)

(99)³ = (100 – 1)³

= (100)³ – 1³ – (3 × 100 × 1) (100 – 1)

= 1000000 – 1 – 300(100 – 1)

= 1000000 – 1 – 30000 + 300

= 970299

(ii) (102)³

Solution:

102 = 100 + 2

Using formula, (x + y)³ = x³ + y³ + 3xy(x + y)

(100 + 2)³ = (100)³ + 2³ + (3 × 100 × 2) (100 + 2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)³

Solution:

998 = 1000 – 2

Using formula, (x – y)³ = x³ – y³ – 3xy(x – y)

(998)³ = (1000 – 2)³

= (1000)³ – 2³ – (3 × 1000 × 2) (1000 – 2)

= 1000000000 – 8 – 6000(1000 –  2)

= 1000000000 – 8 – 6000000 + 12000

= 994011992

Question 8. Factorise each of the following:

(i) 8a³ + b³ + 12a²b + 6ab²

Solution:

8a³ + b³ +12a²b + 6ab² can also be written as (2a)³ + b³ + 3(2a)²b + 3(2a)(b)²

8a³ + b³ + 12a²b + 6ab² = (2a)³ + b³ + 3(2a)²b + 3(2a)(b)²

Formula used, (x + y)³ = x³ + y³ + 3xy(x + y)

= (2a + b)³

= (2a + b) (2a + b) (2a + b)

(ii) 8a³ – b³ – 12a²b + 6ab²

Solution:

8a³ – b³ − 12a²b + 6ab² can also be written as (2a)³– b³ – 3(2a)²b + 3(2a)(b)²

8a³ – b³ − 12a²b + 6ab² = (2a)³ – b³ – 3(2a)²b + 3(2a)(b)²

formula used, (x – y)³ = x³ – y³ – 3xy(x – y)

= (2a – b)³

= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a³ – 135a + 225a²  

Solution:

27 – 125a³ – 135a +225a² can be also written as 3³ – (5a)³ – 3(3)²(5a) + 3(3)(5a)²

27 – 125a³ – 135a + 225a² = 3³ – (5a)³ – 3(3)²(5a) + 3(3)(5a)²

Formula used, (x – y)³ = x³ – y³ – 3xy(x – y)

= (3 – 5a)³

= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a³ – 27b³ – 144a²b + 108ab²

Solution:

64a³ – 27b³ – 144a²b + 108ab² can also be written as (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)²

64a³ – 27b³ – 144a²b + 108ab² = (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)²

Formula used, (x – y)³ = x³ – y³ – 3xy(x – y)

= (4a – 3b)³

= (4a – 3b) (4a – 3b) (4a – 3b)

(v) 7p³ – −  p² + p

Solution:

27p³ –  − () p² + ()p can also be written as (3p)³ –  – 3(3p)²() + 3(3p)()²

27p³ – () − () p² + ()p = (3p)³ – ()³ – 3(3p)²() + 3(3p)()²

Formula used, (x – y)³ = x³ – y³ – 3xy(x – y)

= (3p – )³

= (3p – ) (3p – ) (3p – )

Question 9. Verify:

(i) x³ + y³ = (x + y) (x² – xy + y²)

Solution:

Formula (x + y)³ = x³ + y³ + 3xy(x + y)

x³ + y³ = (x + y)³ – 3xy(x + y)

x³ + y³ = (x + y) [(x + y)² – 3xy]

x³ + y³ = (x + y) [(x² + y² + 2xy) – 3xy]

Therefore, x³ + y³ = (x + y) (x² + y² – xy)

(ii) x³ – y³ = (x – y) (x² + xy + y²)  

Solution:

Formula, (x – y)³ = x³ – y³ – 3xy(x – y)

x³ − y³ = (x – y)³ + 3xy(x – y)

x³− y³ = (x – y) [(x – y)² + 3xy]

 x³ − y³ = (x – y) [(x² + y² – 2xy) + 3xy]

Therefore, x³ + y³ = (x – y) (x² + y² + xy)

Question 10. Factorize each of the following:

(i) 27y³ + 125z³

Solution:

27y³ + 125z³ can also be written as (3y)³ + (5z)³

27y³ + 125z³ = (3y)³ + (5z)³

Formula x³ + y³ = (x + y) (x² – xy + y²)

27y³ + 125z³ = (3y)³ + (5z)³

= (3y + 5z) [(3y)² – (3y)(5z) + (5z)²]

= (3y + 5z) (9y² – 15yz + 25z²)

(ii) 64m³ – 343n³

Solution:

64m³ – 343n³ can also be written as (4m)³ – (7n)³

64m³ – 343n³ = (4m)3 – (7n)³

Formula x³ – y³ = (x – y) (x² + xy + y²)

64m³ – 343n³ = (4m)³ – (7n)³

= (4m – 7n) [(4m)² + (4m)(7n) + (7n)²]

Question 11. Factorise: 27x³ + y³ + z³ – 9xyz  

Solution:

27x³ + y³ + z³ – 9xyz can also be written as (3x)³ + y³ + z³ – 3(3x)(y)(z)

27x³ + y³ + z³ – 9xyz  = (3x)³ + y³ + z³ – 3(3x)(y)(z)

Formula, x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy  – yz – zx)

27x³ + y³ + z³ – 9xyz  = (3x)³ + y³ + z³ – 3(3x)(y)(z)

= (3x + y + z) [(3x)² + y² + z² – 3xy – yz – 3xz]

= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3xz)

Question 12. Verify that: x³ + y³ + z³ – 3xyz  = (1/2) (x + y + z) [(x – y)² + (y – z)² + (z – x)²]

Solution:

Formula, x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² – xy – yz – xz)

Multiplying by 2 and dividing by 2

= (1/2) (x + y + z) [2(x² + y² + z² – xy – yz – xz)]

= (1/2) (x + y + z) (2x² + 2y² + 2z² – 2xy – 2yz – 2xz)

= (1/2) (x + y + z) [(x² + y² − 2xy) + (y² + z² – 2yz) + (x² + z² – 2xz)]

= (1/2) (x + y + z) [(x – y)² + (y – z)² + (z – x)²]

Therefore, x³ + y³ + z³ – 3xyz  = (1/2) (x + y + z) [(x – y)² + (y – z)² + (z – x)²]

Question 13. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.

Solution:

Formula, x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – xz)

Given, (x + y + z) = 0,

Then, x³ + y³ + z³ – 3xyz = (0) (x² + y² + z² – xy – yz – xz)

x³ + y³ + z³ – 3xyz = 0

Therefore, x³ + y³ + z³ = 3xyz

Question 14. Without actually calculating the cubes, find the value of each of the following:

(i) (−12)³ + (7)³ + (5)³

Solution:

Let,

x = −12

y = 7

z = 5

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz.

and we have −12 + 7 + 5 = 0

Therefore, (−12)³ + (7)³ + (5)³ = 3xyz

= 3 × -12 × 7 × 5

= -1260

(ii) (28)³ + (−15)³ + (−13)³

Solution:

Let, 

x = 28

y = −15

z = −13

We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz.

and we have, x + y + z = 28 – 15 – 13 = 0

Therefore, (28)³ + (−15)³ + (−13)³ = 3xyz

= 3 (28) (−15) (−13)

= 16380

Question 15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:  

(i) Area : 25a² – 35a + 12

Solution:

Using splitting the middle term method,

25a² – 35a + 12

25a² – 35a + 12 = 25a² – 15a − 20a + 12

= 5a(5a – 3) – 4(5a – 3)

= (5a – 4) (5a – 3)

Possible expression for length & breadth is  = (5a – 4)  & (5a  – 3)

(ii) Area : 35y² + 13y – 12

Solution:

Using the splitting the middle term method,

35y² + 13y – 12 = 35y² – 15y + 28y – 12

= 5y(7y – 3) + 4(7y – 3)

= (5y + 4) (7y – 3)

Possible expression for length  & breadth is = (5y + 4) & (7y – 3)

Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?  

(i) Volume : 3x² – 12x

Solution:

3x² – 12x can also be written as 3x(x – 4) 

= (3) (x) (x – 4) 

Possible expression for length, breadth & height = 3, x & (x – 4)

(ii) Volume: 12ky² + 8ky – 20k

Solution:

12ky² + 8ky – 20k can also be written as 4k (3y² + 2y – 5)

12ky² + 8ky– 20k = 4k(3y² + 2y – 5)

Using splitting the middle term method.

= 4k (3y2 + 5y – 3y – 5)

= 4k [y(3y + 5) – 1(3y + 5)]

= 4k (3y + 5) (y – 1)

Possible expression for length, breadth & height= 4k, (3y + 5) & (y – 1)

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Also Check:

• NCERT Solutions for Class 9 English
• NCERT Solutions for Class 9 Biology
• NCERT Solutions For Class 11 Biology
• NCERT Solutions for Class 8 to 12

Q1: Why is it important to learn polynomials?

Polynomials are widely employed in many areas of mathematics and other disciplines. Understanding polynomials forms a strong basis for advanced algebra, calculus, and other mathematical concepts that students will come upon in their academic careers.

Q2: What topics are covered in NCERT Answers Class 9 Mathematics Chapter 2 Polynomials?

Polynomials NCERT Solutions for Class 9 Maths– Polynomials covers topics such as the Polynomials and monomials, Types of polynomials, Operations on polynomials, Degree of a polynomial, Zeroes of a polynomial, Factorization of polynomials, Remainder theorem, Algebraic identities and Application problems.

Q3: How can NCERT Solutions for Class 9 Maths Chapter 2 Polynomials help me?

NCERT Solutions for can help you solve the NCERT exercise with Class 9 Maths Chapter 2 – Polynomials out any limitations. If you are stuck on a problem you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

Q4: How many exercises are there in Class 9 Maths Chapter 2 Polynomials?

There are 4 exercises in the Class 9 Maths Chapter 2 – Polynomials which covers all the important topics and sub-topics.

Q5: Where can I find CBSE Class 9 Mathematics Chapter 2 Polynomials solutions?

You can find these NCERT Solutions for Class 9 Maths in this article created by our team of experts at w3wiki