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.
Table of Content
 NCERT Class 9 Maths Chapter 2 Polynomials Topics
 NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.1
 NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.2
 NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.3
 NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.4
 NCERT Solutions for Class 9 Maths Polynomials: Exercise 2.5
 Important Points to Remember
 FAQs on NCERT Solutions for Class 9 Maths Chapter 2
Class 9 Maths NCERT Solutions Chapter 2 Exercises 





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) 4x^{2} – 3x + 7
(ii) y^{2} + √2
(iii) 3√t + t√2
(iv) y + 2/y
(v) x^{10 }+ y^{3 }+ t^{50}
Solution:
(i) The algebraic expression 4x^{2} – 3x + 7 can be written as 4x^{2} – 3x + 7x^{0}
As we can see, all exponents of x are whole numbers,
So, the given expression 4x^{2} – 3x + 7 is polynomial in one variable.
(ii) The algebraic expression y^{2} + √2 can be written as y^{2} + √2y^{0}
As we can see, all exponents of y are whole numbers,
So, the given expression y^{2} + √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 x^{10}+ y^{3}+ t^{50}
As we can see, the expression contains three variables i.e x, y, and t,
So, the given expression x^{10 }+ y^{3 }+ t^{50 }is not a polynomial in one variable.
Question 2. Write the coefficients of x^{2} in each of the following
(i) 2 + x^{2} + x
(ii) 2 – x^{2} + x^{3}
(iii) pi/2 x^{2} + x
(iv) √2x – 1
Solution:
(i) The given algebraic expression is 2 + x^{2} + x
As we can clearly see, the coefficient of x^{2 }is 1.
(ii) The given algebraic expression is 2 – x^{2} + x^{3}
As we can clearly see, the coefficient of x^{2 }is 1.
(iii) The given algebraic expression is pi/2 x^{2} + x
As we can clearly see, the coefficient of x^{2 }is pi/2.
(iv) The given algebraic expression is √2 x — 1
As we can clearly see, the coefficient of x^{2 }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 4x^{35} + 50
A Monomial having degree 100 is 3t^{100}pi
Question 4. Write the degree of each of the following polynomials
(i) 5x^{3 }+ 4x^{2} + 7x
(ii) 4 – y^{2}
(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 5x^{3 }+ 4x^{2} + 7x
As we can clearly see, the highest power of variable x is 3,
So, the degree of given polynomial 5x^{3}+4x^{2} + 7x is 3.
(ii) The given expression is 4 – y^{2}
As we can clearly see, the highest power of variable y is 2,
So, the degree of given polynomial 4 – y^{2} 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 3x^{0}
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) x^{2 }+ x
(ii) x – x^{3}
(iii) y + y^{2 }+ 4
(iv) 1 + x
(v) 3t
(vi) r^{2}
(vii) 7x^{3}
Solution:
(i) Since the degree of given polynomial x^{2 }+ x is 2,
So, it is a Quadratic Polynomial.
(ii) Since the degree of given polynomial x – x^{3} is 3,
So, it is a Cubic Polynomial.
(iii) Since the degree of given polynomial y + y^{2 }+ 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 r^{2} is 2,
So, it is a Quadratic Polynomial.
(vii) Since the degree of given polynomial 7x^{3} 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 − 4x^{2} + 3
(i) x = 0
(ii) x = –1
(iii) x = 2
Solution:
Given equation: 5x − 4x^{2} + 3
Therefore, let f(x) = 5x – 4x^{2 }+ 3
(i) When x = 0
f(0) = 5(0)4(0)^{2}+3
= 3
(ii) When x = 1
f(x) = 5x−4x^{2}+3
f(−1) = 5(−1)−4(−1)^{2}+3
= −5–4+3
= −6
(iii) When x = 2
f(x) = 5x−4x^{2}+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) = y^{2}−y+1
(ii) p(t) = 2+t+2t^{2}−t^{3}
(iii) p(x) = x^{3}
(iv) P(x) = (x−1)(x+1)
Solution:
(i) p(y) = y^{2 }– y + 1
Given equation: p(y) = y^{2}–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) = y^{2}–y+1
(ii) p(t) = 2 + t + 2t^{2 }− t^{3}
Given equation: p(t) = 2+t+2t^{2}−t^{3}
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+2t^{2}−t^{3}
(iii) p(x) = x^{3}
Given equation: p(x) = x^{3}
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)=x^{3}
(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) = x^{2}−1, x=1, −1
(iv) p(x) = (x+1)(x–2), x =−1, 2
(v) p(x) = x^{2}, x = 0
(vi) p(x) = lx+m, x = −m/l
(vii) p(x) = 3x^{2}−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)=x^{2}−1, x=1, −1
Given: p(x)=x^{2}−1 and x=1, −1
Therefore, substituting the value of x in equation p(x), we get.
For x = 1
p(1) = 1^{2}−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) = x^{2}, x = 0
Given: p(x) = x^{2 }and x = 0
Therefore, substituting the value of x in equation p(x), we get.
For, x = 0
p(0) = 0^{2} = 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) = 3x^{2}−1, x = 1/√3 , 2/√3
Given: p(x) = 3x^{2}−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
= 11
= 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 x^{3 }+ 3x^{2 }+ 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
= −pi^{3} + 3pi^{2} − 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 x^{3} − ax^{2} + 6x − a is divided by x – a.
Solution:
Let f(x) = x^{3} − ax^{2} + 6x − a
x − a = 0
∴ x = a
Therefore remainder will be f(x):
f(a) = (a)^{3} − a(a^{2}) + 6(a) − a
= a^{3} − a^{3} + 6a − a
= 5a
Question 3. Check whether 7 + 3x is a factor of 3x^{3 }+ 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 3x^{3 }+ 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) x^{3}+x^{2}+x+1
Solution:
p(x) = x^{3} + x^{2} + 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) x^{4}+x^{3}+x^{2}+x+1
Solution:
p(x) = x^{4}+x^{3}+x^{2}+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) x^{4}+3x^{3}+3x^{2}+x+1
Solution:
p(x) = x^{4}+3x^{3}+3x^{2}+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) x^{3} – x^{2}– (2+√2)x +√2
Solution:
p(x) = x^{3} – x^{2}– (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) = 2x^{3}+x^{2}–2x–1, g(x) = x+1
Solution:
p(x) = 2x^{3}+x^{2}– 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 2x^{3} + x^{2} – 2x – 1
(ii) p(x) = x^{3}+3x^{2}+3x+1, g(x) = x+2
Solution:
p(x) = x^{3}+3x^{2}+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 x^{3} + 3x^{2} +3x + 1
(iii) p(x)=x^{3}– 4x^{2}+x+6, g(x) = x – 3
Solution:
p(x) = x^{3}– 4x^{2}+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) = x^{2}+x+k
Solution:
p(x) = x^{2} + x + k
By Factor Theorem,
As x1 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) = 2x^{2}+kx+√2
Solution:
p(x) = 2x^{2 }+ kx + √2
By Factor Theorem,
As x1 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) = kx^{2}–√2x+1
Solution:
p(x) = kx^{2} – √2x + 1
By Factor Theorem,
As x1 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) = kx^{2}–3x+k
Solution:
p(x) = kx^{2} 3x + k
By Factor Theorem,
As x1 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) 12x^{2}–7x+1
Solution:
p(x) = 12x^{2} – 7x + 1
Using splitting the middle term method,
We need to find a pair of numbers whose sum is 7x
and product is 12x^{2}
7x can be written as the sum of 3x and 4x
12x^{2} can be written as the product of 3x and 4x
=> 12x^{2} – 7x + 1
=> 12x^{2} 3x 4x +1
=> 3x(4x 1) 1(4x 1)
=> (3x – 1)(4x – 1) are the factors of 12x^{2} – 7x + 1
(ii) 2x^{2}+7x+3
Solution:
p(x) = 2x^{2} + 7x + 3
Using splitting the middle term method,
We need to find a pair of numbers whose sum is 7x
and product is 6x^{2}
7x can be written as the sum of 1x and 6x
6x^{2} can be written as the product of 1x and 6x
=> 2x^{2} + 7x + 3
=> 2x^{2} + 1x + 6x + 3
=> 2x(x + 3) + 1(x + 3)
=> (2x + 1)(x + 3) are the factors of 2x^{2} + 7x + 3
(iii) 6x^{2}+5x6
Solution:
p(x) = 6x^{2} + 5x – 6
Using splitting the middle term method,
We need to find a pair of numbers whose sum is 5x
and product is 36x^{2}
5x can be written as the sum of 9x and 4x
36x^{2} can be written as the product of 9x and 4x
=> 6x^{2} + 5x – 6
=> 6x^{2} + 9x – 4x – 6
=> 3x(2x + 3) – 2(2x + 3)
=> (3x – 2)(2x + 3) are the factors of 6x^{2} + 5x – 6
(iv) 3x^{2}–x–4
Solution:
p(x) = 3x^{2} – x – 4
Using splitting the middle term method,
We need to find a pair of numbers whose sum is x
and product is 12x^{2}
x can be written as the sum of 4x and 3x
12x^{2} can be written as the product of 4x and 3x
=> 3x^{2} – x – 4
=> 3x^{2} – 4x + 3x – 4
=> 3x(x + 1) – 4(x + 1)
=> (3x – 4)(x + 1) are the factors of 3x^{2} – x – 4
Question 5. Factorize:
(i) x^{3}–2x^{2}–x+2
Solution:
p(x) = x^{3}– 2x^{2}– 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 x^{3} – 2x^{2} – x + 2
Performing Long Division :
Dividend = Divisor × Quotient + Remainder
=>p(x) = (x – 1)(x^{2} – x – 2)
=> solving (x^{2} – x 2)
=> using Splitting the middle term method
=> x^{2} – 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) x^{3}–3x^{2}–9x–5
Solution:
p(x) = x^{3}–3x^{2}–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)(x^{2 }– 4x – 5)
=> solving (x^{2} – 4x – 5)
=> using splitting the middle term method
=> x^{2} 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) x^{3}+13x^{2}+32x+20
Solution:
p(x) = x^{3}+13x^{2}+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)(x^{2} + 12x + 20)
=> solving x^{2} + 12x + 20
=> using Splitting the middle term method
=> x^{2} + 10x + 2x + 20
=> x(x + 10) + 2(x + 10)
=> (x + 2)(x + 10) are the factors of x^{2} + 12x + 20
=> (x + 1)(x + 2)(x + 10) are the factors of p(x)
(iv) 2y^{3}+y^{2}–2y–1
Solution:
p(y) = 2y^{3}+y^{2}–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)(2y^{2} + 3y + 1)
=> solving 2y^{2} + 3y +1
=> using splitting the middle term method
=> 2y^{2} + 2y +y +1
=> 2y(y + 1)+1(2y + 1)
=>(2y + 1)(2y + 1) are the factors of 2y^{2} + 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) = x^{2} + (a + b)x + ab
[So, a = 4 and b = 10]
(x + 4) (x + 10) = x^{2} + (4 + 10)x + (4 × 10)
= x^{2 }+ 14x + 40
(ii) (x + 8) (x – 10)
Solution:
Using formula, (x + a) (x + b) = x^{2} + (a + b)x + ab
[So, a = 8 and b = −10]
(x + 8) (x – 10) = x^{2} + (8 + (10) )x + (8 × (10))
= x^{2} + (8 – 10) x – 80
= x^{2} − 2x – 80
(iii) (3x + 4) (3x – 5)
Solution:
Using formula, (y + a) (y + b) = y^{2} + (a + b)y + ab
[So, y = 3x, a = 4 and b = −5]
(3x + 4) (3x − 5) = (3x)^{2} + [4 + (5)]3x + 4 × (5)
= 9x^{2} + 3x (4 – 5) – 20
= 9x^{2 }– 3x – 20
(iv) (y^{2} + ) (y^{2} – )
Solution:
Using formula, (a + b) (a – b) = a^{2} – b^{2}
[So, a = y^{2 }and b = ]
(y ^{2} + ) (y^{2} – ) = (y^{2})^{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) = x^{2} + (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) = x^{2} – (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) = a^{2} – b^{2}
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) 9x^{2} + 6xy + y^{2}
Solution:
9x^{2} + 6xy + y^{2} = (3x)^{2} + (2 × 3x × y) + y^{2}
Using formula, a^{2} + 2ab + b^{2} = (a + b)^{2}
Then,
a = 3x
b = y
9x^{2} + 6xy + y^{2} = (3x)^{2} + (2 × 3x × y) + y^{2}
= (3x + y)^{2}
= (3x + y) (3x + y)
(ii) 4y^{2} − 4y + 1
Solution:
4y^{2 }− 4y + 1 = (2y)^{2} – (2 × 2y × 1) + 1
Using formula, a^{2} – 2ab + b^{2} = (a – b)^{2}
Then,
a = 2y
b = 1
= (2y – 1)^{2}
= (2y – 1) (2y – 1)
(iii) x^{2} –
Solution:
x^{2} – = x^{2} –
Using formula, a^{2} – b^{2} = (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} = x^{2 }+ y^{2} + z^{2} + 2xy + 2yz + 2zx
Then,
x = x
y = 2y
z = 4z
(x + 2y + 4z)^{2} = x^{2} + (2y)^{2} + (4z)^{2} + (2 × x × 2y) + (2 × 2y × 4z) + (2 × 4z × x)
= x^{2} + 4y^{2} + 16z^{2} + 4xy + 16yz + 8xz
(ii) (2x − y + z)^{2}
Solution:
Using formula, (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx
Then,
x = 2x
y = −y
z = z
(2x − y + z)^{2} = (2x)^{2} + (−y)^{2} + z^{2} + (2 × 2x × −y) + (2 × −y × z) + (2 × z × 2x)
= 4x^{2} + y^{2} + z^{2} – 4xy – 2yz + 4xz
(iii) (−2x + 3y + 2z)^{2}
Solution:
Using formula, (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 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)
= 4x^{2} + 9y^{2} + 4z^{2} – 12xy + 12yz– 8xz
(iv) (3a – 7b – c)^{2}
Solution:
Using formula, (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 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)
= 9a^{2} + 49b^{2} + c^{2} – 42ab + 14bc – 6ca
(v) (–2x + 5y – 3z)^{2}
Solution:
Using formula, (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 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)
= 4x^{2} + 25y^{2} + 9z^{2} – 20xy – 30yz + 12zx
(vi) (a – b + 1)^{2}
Solution:
Using formula, (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx
Then,
x = a
y = b
z = 1
(a – ()b + 1)^{2} = [a]^{2} + [b]^{2} + 1^{2} + [2 x }a x b] + [2 xb x 1] + [2 x 1 x a]
= a^{2} + b^{2} + 1 – ab – b + a
Question 5. Factorize:
(i) 4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz
Solution:
Using formula, (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx
Then, x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx = (x + y + z)^{2}
4x^{2} + 9y^{2} + 16z^{2} + 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) 2x^{2} + y^{2} + 8z^{2} – 2√2xy + 4√2yz – 8xz
Solution:
Using formula, (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx
Then, x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx = (x + y + z)^{2}
2x^{2} + y^{2} + 8z^{2} – 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} = x^{3} + y^{3} + 3xy(x + y)
(2x + 1)^{3}= (2x)^{3} + 1^{3} + (3 × 2x ×1) (2x + 1)
= 8x^{3} + 1 + 6x(2x + 1)
= 8x^{3} + 12x^{2} + 6x + 1
(ii) (2a − 3b)^{3}
Solution:
Using formula, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
(2a − 3b)^{3} = (2a)^{3} − (3b)^{3} – (3 × 2a × 3b) (2a – 3b)
= 8a^{3} – 27b^{3} – 18ab(2a – 3b)
= 8a^{3} – 27b^{3} – 36a^{2}b + 54ab^{2}
(iii) (x + 1)^{3}
Solution:
Using formula, (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)
(x+ 1)^{3} = (x)^{3} + 1^{3} + (3 × x × 1) (x + 1)
= x^{3} + 1 + x^{2} + x
=
(iv) (x − y)^{3}
Solution:
Using formula, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
(x − y)^{3} = x^{3} − [y]^{3} – 3(x) y[x − y]
= x^{3} –y^{3} – 2x^{2}y + xy^{2}
Question 7. Evaluate the following using suitable identities:
(i) (99)^{3}
Solution:
99 = 100 – 1
Using formula, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
(99)^{3} = (100 – 1)^{3}
= (100)^{3} – 1^{3} – (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} = x^{3} + y^{3} + 3xy(x + y)
(100 + 2)^{3} = (100)^{3} + 2^{3} + (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} = x^{3} – y^{3} – 3xy(x – y)
(998)^{3} = (1000 – 2)^{3}
= (1000)^{3} – 2^{3} – (3 × 1000 × 2) (1000 – 2)
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 + 12000
= 994011992
Question 8. Factorise each of the following:
(i) 8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}
Solution:
8a^{3} + b^{3} +12a^{2}b + 6ab^{2} can also be written as (2a)^{3} + b^{3} + 3(2a)^{2}b + 3(2a)(b)^{2}
8a^{3} + b^{3} + 12a^{2}b + 6ab^{2} = (2a)^{3} + b^{3} + 3(2a)^{2}b + 3(2a)(b)^{2}
Formula used, (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)
= (2a + b)^{3}
= (2a + b) (2a + b) (2a + b)
(ii) 8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}
Solution:
8a^{3} – b^{3} − 12a^{2}b + 6ab^{2} can also be written as (2a)^{3}– b^{3} – 3(2a)^{2}b + 3(2a)(b)^{2}
8a^{3} – b^{3} − 12a^{2}b + 6ab^{2} = (2a)^{3} – b^{3} – 3(2a)^{2}b + 3(2a)(b)^{2}
formula used, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
= (2a – b)^{3}
= (2a – b) (2a – b) (2a – b)
(iii) 27 – 125a^{3} – 135a + 225a^{2}
Solution:
27 – 125a^{3} – 135a +225a^{2} can be also written as 3^{3} – (5a)^{3} – 3(3)^{2}(5a) + 3(3)(5a)^{2}
27 – 125a^{3} – 135a + 225a^{2} = 3^{3} – (5a)^{3} – 3(3)^{2}(5a) + 3(3)(5a)^{2}
Formula used, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
= (3 – 5a)^{3}
= (3 – 5a) (3 – 5a) (3 – 5a)
(iv) 64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}
Solution:
64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2} can also be written as (4a)^{3} – (3b)^{3} – 3(4a)^{2}(3b) + 3(4a)(3b)^{2}
64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2} = (4a)^{3} – (3b)^{3} – 3(4a)^{2}(3b) + 3(4a)(3b)^{2}
Formula used, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
= (4a – 3b)^{3}
= (4a – 3b) (4a – 3b) (4a – 3b)
(v) 7p^{3} – − p^{2} + p
Solution:
27p^{3} – − () p^{2} + ()p can also be written as (3p)^{3} – – 3(3p)^{2}() + 3(3p)()^{2}
27p^{3} – () − () p^{2} + ()p = (3p)^{3} – ()^{3} – 3(3p)^{2}() + 3(3p)()^{2}
Formula used, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
= (3p – )^{3}
= (3p – ) (3p – ) (3p – )
Question 9. Verify:
(i) x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})
Solution:
Formula (x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)
x^{3} + y^{3} = (x + y)^{3} – 3xy(x + y)
x^{3} + y^{3} = (x + y) [(x + y)^{2} – 3xy]
x^{3} + y^{3} = (x + y) [(x^{2} + y^{2} + 2xy) – 3xy]
Therefore, x^{3} + y^{3} = (x + y) (x^{2} + y^{2 }– xy)
(ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})
Solution:
Formula, (x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
x^{3} − y^{3} = (x – y)^{3} + 3xy(x – y)
x^{3}− y^{3} = (x – y) [(x – y)^{2} + 3xy]
x^{3} − y^{3} = (x – y) [(x^{2 }+ y^{2 }– 2xy) + 3xy]
Therefore, x^{3} + y^{3} = (x – y) (x^{2} + y^{2} + xy)
Question 10. Factorize each of the following:
(i) 27y^{3} + 125z^{3}
Solution:
27y^{3} + 125z^{3} can also be written as (3y)^{3} + (5z)^{3}
27y^{3} + 125z^{3} = (3y)^{3} + (5z)^{3}
Formula x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})
27y^{3} + 125z^{3} = (3y)^{3} + (5z)^{3}
= (3y + 5z) [(3y)^{2} – (3y)(5z) + (5z)^{2}]
= (3y + 5z) (9y^{2} – 15yz + 25z^{2})
(ii) 64m^{3} – 343n^{3}
Solution:
64m^{3} – 343n^{3} can also be written as (4m)^{3} – (7n)^{3}
64m^{3} – 343n^{3} = (4m)3 – (7n)^{3}
Formula x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})
64m^{3} – 343n^{3} = (4m)^{3} – (7n)^{3}
= (4m – 7n) [(4m)^{2} + (4m)(7n) + (7n)^{2}]
Question 11. Factorise: 27x^{3} + y^{3} + z^{3} – 9xyz
Solution:
27x^{3} + y^{3} + z^{3} – 9xyz can also be written as (3x)^{3} + y^{3} + z^{3} – 3(3x)(y)(z)
27x^{3} + y^{3} + z^{3} – 9xyz = (3x)^{3} + y^{3} + z^{3} – 3(3x)(y)(z)
Formula, x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)
27x^{3} + y^{3} + z^{3} – 9xyz = (3x)^{3} + y^{3} + z^{3} – 3(3x)(y)(z)
= (3x + y + z) [(3x)^{2} + y^{2} + z^{2} – 3xy – yz – 3xz]
= (3x + y + z) (9x^{2} + y^{2} + z^{2} – 3xy – yz – 3xz)
Question 12. Verify that: x^{3} + y^{3} + z^{3} – 3xyz = (1/2) (x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]
Solution:
Formula, x^{3} + y^{3} + z^{3 }− 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – xz)
Multiplying by 2 and dividing by 2
= (1/2) (x + y + z) [2(x^{2} + y^{2} + z^{2} – xy – yz – xz)]
= (1/2) (x + y + z) (2x^{2} + 2y^{2} + 2z^{2} – 2xy – 2yz – 2xz)
= (1/2) (x + y + z) [(x^{2} + y^{2} − 2xy) + (y^{2} + z^{2} – 2yz) + (x^{2} + z^{2} – 2xz)]
= (1/2) (x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]
Therefore, x^{3} + y^{3} + z^{3} – 3xyz = (1/2) (x + y + z) [(x – y)^{2} + (y – z)^{2} + (z – x)^{2}]
Question 13. If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3xyz.
Solution:
Formula, x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – xz)
Given, (x + y + z) = 0,
Then, x^{3} + y^{3} + z^{3} – 3xyz = (0) (x^{2} + y^{2} + z^{2} – xy – yz – xz)
x^{3} + y^{3} + z^{3} – 3xyz = 0
Therefore, x^{3} + y^{3} + z^{3} = 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 x^{3} + y^{3} + z^{3} = 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 x^{3} + y^{3} + z^{3} = 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 : 25a^{2} – 35a + 12
Solution:
Using splitting the middle term method,
25a^{2} – 35a + 12
25a^{2} – 35a + 12 = 25a^{2} – 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 : 35y^{2} + 13y – 12
Solution:
Using the splitting the middle term method,
35y^{2} + 13y – 12 = 35y^{2} – 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 : 3x^{2} – 12x
Solution:
3x^{2 }– 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: 12ky^{2} + 8ky – 20k
Solution:
12ky^{2} + 8ky – 20k can also be written as 4k (3y^{2} + 2y – 5)
12ky^{2} + 8ky– 20k = 4k(3y^{2} + 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 stepbystep format for better understanding.
Also Check:
 NCERT Solutions for Class 9 English
 NCERT Solutions for Class 9 Biology
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 NCERT Solutions for Class 8 to 12
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 subtopics.
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