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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Last Updated : 08 Nov, 2023
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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.

Class 9 Maths NCERT Solutions Chapter 2 Exercises

  • NCERT Maths Solutions Class 9 Exercise 2.5 Set 1Set 2 – 16 Questions (9 Short Answers, 5 Long Answers, 2 Very Long Answers)

NCERT Class 9 Maths Chapter 2 Polynomials Topics

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.

NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.1

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 t1/2 + √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.

NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.2

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

NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.3

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

NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.4

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 :

Quotient : x2 – x – 2 , Remainder : 0

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 :

Quotient : x2 – 4x – 5, Remainder : 0

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 :

Quotient : x2 + 12x + 20, Remainder : 0

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 :

Quotient : 2y2 + 3y + 1, Remainder : 0

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 for Class 9 Maths Polynomials: Exercise 2.5

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\frac{3}{2}                 ) (y2 – \frac{3}{2}                 )

Solution:

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

[So, a = y2 and b = \frac{3}{2}                 ]

(y 2\frac{3}{2}                 ) (y2 – \frac{3}{2}                 ) = (y2)2 – (\frac{3}{2}                 )^2

= y 4 – \frac{9}{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 – \frac{y^2}{100}

Solution:

x2 – \frac{y^2}{100}                  = x2 – (\frac{y}{10})^2

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

Then, 

a = x

b = \frac{y}{10}

= (x – \frac{y}{10}                 ) (x + \frac{y}{10}                 )

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) (\frac{1}{4}                 a – \frac{1}{2}                 b + 1)2

Solution:

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

Then, 

x = \frac{1}{4}                 a

y = \frac{-1}{2}                 b

z = 1

(\frac{1}{4}                 a – (\frac{-1}{2}                 )b + 1)2 = [\frac{1}{4}                 a]2 + [\frac{-1}{2}                 b]2 + 12 + [2 x \frac{1}{4}                 }a x \frac{-1}{2}                 b] + [2 x\frac{-1}{2}                 b x 1] + [2 x 1 x  \frac{1}{4}                 a]

\frac{1}{16}                 a2\frac{1}{4}                 b2 + 1 – \frac{1}{4}                 ab – b + \frac{1}{2}                 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) (\frac{3}{2}                 x + 1)3

Solution:

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

(\frac{3}{2}                 x+ 1)3 = (\frac{3}{2}                 x)3 + 13 + (3 × \frac{3}{2}                 x × 1) (\frac{3}{2}                 x + 1)

\frac{27}{8}                 x3 + 1 + \frac{27}{4}                 x2\frac{9}{2}                 x

\frac{27}{8}x^3  + \frac{27}{4}x^2 + \frac{9}{2}x +1

(iv) (x − \frac{2}{3}                 y)3

Solution:

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

(x − \frac{2}{3}                 y)3 = x3 − [\frac{2}{3}                 y]3 – 3(x) \frac{2}{3}                 y[x − \frac{2}{3}                 y]

= x3\frac{8}{27}                 y3 – 2x2y + \frac{4}{3}                 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\frac{1}{216}                  − \frac{9}{2}                  p2\frac{1}{4}                 p

Solution:

27p3 – \frac{1}{216}                  − (\frac{9}{2}                 ) p2 + (\frac{1}{4}            )p can also be written as (3p)3 – (\frac{1}{6})^3             – 3(3p)2(\frac{1}{6}                 ) + 3(3p)(\frac{1}{6}                 )2

27p3 – (\frac{1}{216}            ) − (\frac{9}{2}            ) p2 + (\frac{1}{4}            )p = (3p)3 – (\frac{1}{6}            )3 – 3(3p)2(\frac{1}{6}            ) + 3(3p)(\frac{1}{6}            )2

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

= (3p – \frac{1}{6}            )3

= (3p – \frac{1}{6}            ) (3p – \frac{1}{6}            ) (3p – \frac{1}{6}            )

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)

Important Points to Remember

  • NCERT Solutions for Class 9 Maths are created by a team of professionals a GfG, with the intention to benefit students.
  • These solutions are very accurate and comprehensive, which can help students prepare for any academic as well as competitive exam.
  • All the solutions provided are in a step-by-step format for better understanding.

Also Check:

FAQs on NCERT Solutions for Class 9 Maths Chapter 2

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 GeeksforGeeks



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