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 chapterβs 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 |
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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 β 3x + 7
(ii) y2 + β2
(iii) 3βt + tβ2
(iv) y + 2/y
(v) x10 + y3 + t50
Solution:
(i) The algebraic expression 4x2 β 3x + 7 can be written as 4x2 β 3x + 7x0
As we can see, all exponents of x are whole numbers,
So, the given expression 4x2 β 3x + 7 is polynomial in one variable.
(ii) The algebraic expression y2 + β2 can be written as y2 + β2y0
As we can see, all exponents of y are whole numbers,
So, the given expression y2 + β2 is polynomial in one variable.
(iii) The algebraic expression 3 βt + tβ2 can be written as 3 t1/2 + β2.t
As we can see, one exponent of t is 1/2, which is not a whole number,
So, the given expression 3 βt + tβ2 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 β x2 + x3
(iii) pi/2 x2 + x
(iv) β2x β 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 β 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 β2 x β 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 β y2
(iii) 5t β β7
(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 β y2
As we can clearly see, the highest power of variable y is 2,
So, the degree of given polynomial 4 β y2 is 2.
(iii) The given expression is 5t β β7
As we can clearly see, the highest power of variable t is 1,
So, the degree of given polynomial 5t β β7 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 β 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 β 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 β 4x2 + 3
(i) x = 0
(ii) x = β1
(iii) x = 2
Solution:
Given equation: 5x β 4x2 + 3
Therefore, let f(x) = 5x β 4x2 + 3
(i) When x = 0
f(0) = 5(0)-4(0)2+3
= 3
(ii) When x = -1
f(x) = 5xβ4x2+3
f(β1) = 5(β1)β4(β1)2+3
= β5β4+3
= β6
(iii) When x = 2
f(x) = 5xβ4x2+3
f(2) = 5(2)β4(2)2+3
= 10β16+3
= β3
Question 2: Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2βy+1
(ii) p(t) = 2+t+2t2βt3
(iii) p(x) = x3
(iv) P(x) = (xβ1)(x+1)
Solution:
(i) p(y) = y2 β y + 1
Given equation: p(y) = y2βy+1
Therefore, p(0) = (0)2β(0)+1 = 1
p(1) = (1)2β(1)+1 = 1
p(2) = (2)2β(2)+1 = 3
Hence, p(0) = 1, p(1) = 1, p(2) = 3, for the equation p(y) = y2βy+1
(ii) p(t) = 2 + t + 2t2 β t3
Given equation: p(t) = 2+t+2t2βt3
Therefore, p(0) = 2+0+2(0)2β(0)3 = 2
p(1) = 2+1+2(1)2β(1)3 = 2+1+2β1 = 4
p(2) = 2+2+2(2)2β(2)3 = 2+2+8β8 = 4
Hence, p(0) = 2,p(1) =4 , p(2) = 4, for the equation p(t)=2+t+2t2βt3
(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) = (xβ1)(x+1)
Given equation: p(x) = (xβ1)(x+1)
Therefore, p(0) = (0β1)(0+1) = (β1)(1) = β1
p(1) = (1β1)(1+1) = 0(2) = 0
p(2) = (2β1)(2+1) = 1(3) = 3
Hence, p(0)= 1, p(1) = 0, p(2) = 3, for the equation p(x) = (xβ1)(x+1)
Question 3: Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x+1, x=β1/3
(ii) p(x) = 5xβΟ, x = 4/5
(iii) p(x) = x2β1, x=1, β1
(iv) p(x) = (x+1)(xβ2), x =β1, 2
(v) p(x) = x2, x = 0
(vi) p(x) = lx+m, x = βm/l
(vii) p(x) = 3x2β1, x = -1/β3 , 2/β3
(viii) p(x) = 2x+1, x = 1/2
Solution:
(i) p(x)=3x+1, x=β1/3
Given: p(x)=3x+1 and x=β1/3
Therefore, substituting the value of x in equation p(x), we get.
For, x = -1/3
p(β1/3) = 3(-1/3)+1
= β1+1
= 0
Hence, p(x) of -1/3 = 0
(ii) p(x)=5xβΟ, x = 4/5
Given: p(x)=5xβΟ 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)βΟ
= 4βΟ
Hence, p(x) of 4/5 β 0
(iii) p(x)=x2β1, x=1, β1
Given: p(x)=x2β1 and x=1, β1
Therefore, substituting the value of x in equation p(x), we get.
For x = 1
p(1) = 12β1
=1β1
= 0
For, x = -1
p(β1) = (-1)2β1
= 1β1
= 0
Hence, p(x) of 1 and -1 = 0
(iv) p(x) = (x+1)(xβ2), x =β1, 2
Given: p(x) = (x+1)(xβ2) and x =β1, 2
Therefore, substituting the value of x in equation p(x), we get.
For, x = β1
p(β1) = (β1+1)(β1β2)
= (0)(β3)
= 0
For, x = 2
p(2) = (2+1)(2β2)
= (3)(0)
= 0
Hence, p(x) of β1, 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 = βm/l
Given: p(x) = lx+m and x = βm/l
Therefore, substituting the value of x in equation p(x), we get.
For, x = βm/l
p(-m/l)= l(-m/l)+m
= βm+m
= 0
Hence, p(x) of -m/l = 0
(vii) p(x) = 3x2β1, x = -1/β3 , 2/β3
Given: p(x) = 3x2β1 and x = -1/β3 , 2/β3
Therefore, substituting the value of x in equation p(x), we get.
For, x = -1/β3
p(-1/β3) = 3(-1/β3)2 -1
= 3(1/3)-1
= 1-1
= 0
For, x = 2/β3
p(2/β3) = 3(2/β3)2 -1
= 3(4/3)-1
= 4β1
=3 β 0
Hence, p(x) of -1/β3 = 0
but, p(x) of 2/β3 β 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
= 2β 0
Hence, p(x) of 1/2 β 0
Question 4: Find the zero of the polynomials in each of the following cases:
(i) p(x) = x+5
(ii) p(x) = xβ5
(iii) p(x) = 2x+5
(iv) p(x) = 3xβ2
(vii) p(x) = cx+d, c β 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 = β5
Therefore, the zero of the polynomial p(x) = x+5 is when x = -5
(ii) p(x) = xβ5
Given: p(x) = xβ5
p(x) = xβ5
xβ5 = 0
x = 5
Therefore, the zero of the polynomial p(x) = xβ5 is when x = 5
(iii) p(x) = 2x+5
Given: p(x) = 2x+5
p(x) = 2x+5
2x+5 = 0
2x = β5
x = -5/2
Therefore, the zero of the polynomial p(x) = 2x+5 is when x = -5/2
(iv) p(x) = 3xβ2
Given: p(x) = 3xβ2
p(x) = 3xβ2
3xβ2 = 0
3x = 2
x = 2/3
Therefore, the zero of the polynomial p(x) = 3xβ2 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, aβ 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 β 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 = β1
Therefore remainder will be f(x):
f(β1) = (β1)3 + 3(β1)2 + 3(β1) + 1
= β1 + 3 β 3 + 1
= 0
(ii) x β 1/2
Solution:
x β 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 = βpi
Therefore remainder will be f(x):
f(βpi) = (βpi)3 + 3(βpi)2 + 3(βpi) + 1
= βpi3 + 3pi2 β 3pi + 1
(v) 5 + 2x
Solution:
5 + 2x = 0
2x = β5
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) β (15/2) + 1
= -27/8
Question 2. Find the remainder when x3 β ax2 + 6x β a is divided by x β a.
Solution:
Let f(x) = x3 β ax2 + 6x β a
x β a = 0
β΄ x = a
Therefore remainder will be f(x):
f(a) = (a)3 β a(a2) + 6(a) β a
= a3 β a3 + 6a β a
= 5a
Question 3. Check whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
7 + 3x = 0
3x = β7
x = -7/3
Therefore remainder will be f(x):
f(-7/3) = 3(-7/3)3 + 7(-7/3)
= β (343/9) + (-49/3)
= (-343- (49) * 3)/9
= (-343 β 147)/9
= β 490/9 β 0
β΄ 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
=> β 1 + 1 β 1 + 1 -1
=> -1
=> -1 β 0
As p(-1) β 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 β 3 + 3 β 1 + 1
=> -1
=> -1 β 0
As p(-1) β 0 so (x + 1) is not a factor of p(x).
(iv) x3 β x2β (2+β2)x +β2
Solution:
p(x) = x3 β x2β (2+β2)x +β2
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 β (-1)2β (2+β2)(-1) +β2
=> -1 β 1 + 2 + β2 + β2
=> 2β2
=> 2β2 β 0
As p(-1) β 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+x2β2xβ1, g(x) = x+1
Solution:
p(x) = 2x3+x2β 2xβ1
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 β 2(-1) -1
=> -2 + 1 + 2 β 1
=> 0
As p(-1) = 0 therefore (x + 1) is a factor of 2x3 + x2 β 2x β 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 β 6 + 1
=> -1
=> -1 β 0
As p(-2) β 0 therefore (x + 2) is not a factor of x3 + 3x2 +3x + 1
(iii) p(x)=x3β 4x2+x+6, g(x) = x β 3
Solution:
p(x) = x3β 4x2+x+6
g(x) = x β 3
By Factor Theorem we know that if x β 3 is a factor of p(x)
Then value of p(3) should be 0
Checking,
=> p(3) = (3)3 β 4(3)2 + 3 + 6
=> 27 β 36 + 3 + 6
=> 0
As p(3)=0 so (x β 3) is a factor of p(x).
Question 3. Find the value of k, if xβ1 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+β2
Solution:
p(x) = 2x2 + kx + β2
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) + β2
=> 2 + k + β2 = 0
=> 2 + β2 + k = 0
=> k = β (2 + β2)
(iii) p(x) = kx2ββ2x+1
Solution:
p(x) = kx2 β β2x + 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 β β2(1) + 1
=> k β β2 + 1 = 0
=> k = β2 β 1
(iv) p(x) = kx2β3x+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 β 3(1) + k
=> k β 3 + k = 0
=> 2k β 3 = 0
=> k = 3/2
Question 4. Factorize:
(i) 12x2β7x+1
Solution:
p(x) = 12x2 β 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 β 7x + 1
=> 12x2 -3x -4x +1
=> 3x(4x -1) -1(4x -1)
=> (3x β 1)(4x β 1) are the factors of 12x2 β 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 β 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 β 6
=> 6x2 + 9x β 4x β 6
=> 3x(2x + 3) β 2(2x + 3)
=> (3x β 2)(2x + 3) are the factors of 6x2 + 5x β 6
(iv) 3x2βxβ4
Solution:
p(x) = 3x2 β x β 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 β x β 4
=> 3x2 β 4x + 3x β 4
=> 3x(x + 1) β 4(x + 1)
=> (3x β 4)(x + 1) are the factors of 3x2 β x β 4
Question 5. Factorize:
(i) x3β2x2βx+2
Solution:
p(x) = x3β 2x2β x + 2
Factors of 2 are Β±1 and Β± 2
Using Hit and Trial Method
p(1) = (1)3 β 2(1)2 β (1) + 2
p(1) = 1 β 2 β 1 + 2
p(1) = 0
Therefore, (x β 1) is a factor of x3 β 2x2 β x + 2
Performing Long Division :
Dividend = Divisor Γ Quotient + Remainder
=>p(x) = (x β 1)(x2 β x β 2)
=> solving (x2 β x -2)
=> using Splitting the middle term method
=> x2 β 2x + x β 2
=> x(x β 2) + 1(x β 2)
=> (x + 1)(x β 2)
=>(x β 1)(x + 1)(x β 2) are the factors of p(x)
(ii) x3β3x2β9xβ5
Solution:
p(x) = x3β3x2β9xβ5
Factors of -5 are Β±1 and Β± 5
Using Hit and Trial Method
let x = 1
p(1) = (1)3 β 3(1)2 β 9(1) β 5
p(1) = 1 β 3 β 9 -5
p(1) = -16
p(1) β 0
let x = -1
p(-1) = (-1)3 β 3(-1)2 β 9(-1) β 5
p(-1) = -1 β 3 + 9 β 5
p(-1) = -9 + 9
p(-1) = 0
Therefore, (x + 1) is a factor of p(x)
Performing Long Division :
Dividend = Divisor Γ Quotient + Remainder
=>p(x) = (x + 1)(x2 β 4x β 5)
=> solving (x2 β 4x β 5)
=> using splitting the middle term method
=> x2 -5x + x β 5
=> x(x β 5) + 1(x β 5)
=> (x + 1)(x β 5)
=>(x + 1)(x + 1)(x β 5) are the factors of p(x)
(iii) x3+13x2+32x+20
Solution:
p(x) = x3+13x2+32x+20
=> Factors of 20 are Β±1, Β±2, Β±4, Β±5, Β±10 and Β±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) β 0
let x = -1
p(-1) = (-1)3 + 13(-1)2 + 32(-1) + 20
p(-1) = -1 + 13 β 32 + 20
p(-1) = -33 + 33
p(-1) = 0
Therefore, (x + 1) is a factor of p(x)
Performing Long Division :
Dividend = Divisor Γ 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+y2β2yβ1
Solution:
p(y) = 2y3+y2β2yβ1
=> Factors of -1 are Β±1
Using Hit and Trial Method
let x = 1
p(1) = 2(1)3 + (1)2 β 2(1) β 1
p(1) = 2 + 1 -2 β 1
p(1) = 0
Therefore, (y β 1) is a factor of p(y)
Performing Long Division :
Dividend = Divisor Γ Quotient + Remainder
=> p(y) = (y β 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 β 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 Γ 10)
= x2 + 14x + 40
(ii) (x + 8) (x β 10)
Solution:
Using formula, (x + a) (x + b) = x2 + (a + b)x + ab
[So, a = 8 and b = β10]
(x + 8) (x β 10) = x2 + (8 + (-10) )x + (8 Γ (-10))
= x2 + (8 β 10) x β 80
= x2 β 2x β 80
(iii) (3x + 4) (3x β 5)
Solution:
Using formula, (y + a) (y + b) = y2 + (a + b)y + ab
[So, y = 3x, a = 4 and b = β5]
(3x + 4) (3x β 5) = (3x)2 + [4 + (-5)]3x + 4 Γ (-5)
= 9x2 + 3x (4 β 5) β 20
= 9x2 β 3x β 20
(iv) (y2 + ) (y2 β )
Solution:
Using formula, (a + b) (a β b) = a2 β b2
[So, a = y2 and b = ]
(y 2 + ) (y2 β ) = (y2)2 β ()^2
= y 4 β
Question 2. Evaluate the following products without multiplying directly:
(i) 103 Γ 107
Solution:
103 Γ 107 = (100 + 3) Γ (100 + 7)
Using formula, (x + a) (x + b) = x2 + (a + b)x + ab
Then,
x = 100
a = 3
b = 7
So, 103 Γ 107 = (100 + 3) Γ (100 + 7)
= (100)2 + (3 + 7)100 + (3 Γ 7)
= 10000 + 1000 + 21
= 11021
(ii) 95 Γ 96
Solution:
95 Γ 96 = (100 β 5) Γ (100 β 4)
Using formula, (x β a) (x β b) = x2 β (a + b)x + ab
Then, According to the identity
x = 100
a = 5
b = 4
So, 95 Γ 96 = (100 β 5) Γ (100 β 4)
= (100)2 β 100 (5+4) + (5 Γ 4)
= 10000 β 900 + 20
= 9120
(iii) 104 Γ 96
Solution:
104 Γ 96 = (100 + 4) Γ (100 β 4)
Using formula, (a + b) (a β b) = a2 β b2
Then,
a = 100
b = 4
So, 104 Γ 96 = (100 + 4) Γ (100 β 4)
= (100)2 β (4)2
= 10000 β 16
= 9984
Question 3. Factorize the following using appropriate identities:
(i) 9x2 + 6xy + y2
Solution:
9x2 + 6xy + y2 = (3x)2 + (2 Γ 3x Γ y) + y2
Using formula, a2 + 2ab + b2 = (a + b)2
Then,
a = 3x
b = y
9x2 + 6xy + y2 = (3x)2 + (2 Γ 3x Γ y) + y2
= (3x + y)2
= (3x + y) (3x + y)
(ii) 4y2 β 4y + 1
Solution:
4y2 β 4y + 1 = (2y)2 β (2 Γ 2y Γ 1) + 1
Using formula, a2 β 2ab + b2 = (a β b)2
Then,
a = 2y
b = 1
= (2y β 1)2
= (2y β 1) (2y β 1)
(iii) x2 β
Solution:
x2 β = x2 β
Using formula, a2 β b2 = (a β b) (a + b)
Then,
a = x
b =
= (x β ) (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 Γ x Γ 2y) + (2 Γ 2y Γ 4z) + (2 Γ 4z Γ x)
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz
(ii) (2x β y + z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = 2x
y = βy
z = z
(2x β y + z)2 = (2x)2 + (βy)2 + z2 + (2 Γ 2x Γ βy) + (2 Γ βy Γ z) + (2 Γ z Γ 2x)
= 4x2 + y2 + z2 β 4xy β 2yz + 4xz
(iii) (β2x + 3y + 2z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = β2x
y = 3y
z = 2z
(β2x + 3y + 2z)2 = (β2x)2 + (3y)2 + (2z)2 + (2 Γβ2x Γ 3y) + (2 Γ3y Γ 2z) + (2 Γ2z Γ β2x)
= 4x2 + 9y2 + 4z2 β 12xy + 12yzβ 8xz
(iv) (3a β 7b β c)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = 3a
y = β 7b
z = β c
(3a β 7b β c)2 = (3a)2 + (β 7b)2 + (β c)2 + (2 Γ 3a Γ β 7b) + (2 Γ β7b Γ βc) + (2 Γ βc Γ 3a)
= 9a2 + 49b2 + c2 β 42ab + 14bc β 6ca
(v) (β2x + 5y β 3z)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = β2x
y = 5y
z = β 3z
(β2x + 5y β 3z)2 = (β2x)2 + (5y)2 + (β3z)2 + (2 Γ β2x Γ 5y) + (2 Γ 5y Γ β 3z) + (2 Γ β3z Γ β2x)
= 4x2 + 25y2 + 9z2 β 20xy β 30yz + 12zx
(vi) (a β b + 1)2
Solution:
Using formula, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
Then,
x = a
y = b
z = 1
(a β ()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 β ab β b + a
Question 5. Factorize:
(i) 4x2 + 9y2 + 16z2 + 12xy β 24yz β 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 β 24yz β 16xz = (2x)2 + (3y)2 + (β4z)2 + (2 Γ 2x Γ 3y) + (2 Γ 3y Γ β4z) + (2 Γ β4z Γ 2x)
= (2x + 3y β 4z)2
= (2x + 3y β 4z) (2x + 3y β 4z)
(ii) 2x2 + y2 + 8z2 β 2β2xy + 4β2yz β 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 β 2β2xy + 4β2yz β 8xz
= (-β2x)2 + (y)2 + (2β2z)2 + (2 Γ -β2x Γ y) + (2 Γ y Γ 2β2z) + (2 Γ 2β2 Γ ββ2x)
= (ββ2x + y + 2β2z)2
= (ββ2x + y + 2β2z) (ββ2x + y + 2β2z)
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 Γ 2x Γ1) (2x + 1)
= 8x3 + 1 + 6x(2x + 1)
= 8x3 + 12x2 + 6x + 1
(ii) (2a β 3b)3
Solution:
Using formula, (x β y)3 = x3 β y3 β 3xy(x β y)
(2a β 3b)3 = (2a)3 β (3b)3 β (3 Γ 2a Γ 3b) (2a β 3b)
= 8a3 β 27b3 β 18ab(2a β 3b)
= 8a3 β 27b3 β 36a2b + 54ab2
(iii) (x + 1)3
Solution:
Using formula, (x + y)3 = x3 + y3 + 3xy(x + y)
(x+ 1)3 = (x)3 + 13 + (3 Γ x Γ 1) (x + 1)
= x3 + 1 + x2 + x
=
(iv) (x β y)3
Solution:
Using formula, (x β y)3 = x3 β y3 β 3xy(x β y)
(x β y)3 = x3 β [y]3 β 3(x) y[x β y]
= x3 βy3 β 2x2y + xy2
Question 7. Evaluate the following using suitable identities:
(i) (99)3
Solution:
99 = 100 β 1
Using formula, (x β y)3 = x3 β y3 β 3xy(x β y)
(99)3 = (100 β 1)3
= (100)3 β 13 β (3 Γ 100 Γ 1) (100 β 1)
= 1000000 β 1 β 300(100 β 1)
= 1000000 β 1 β 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 Γ 100 Γ 2) (100 + 2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii) (998)3
Solution:
998 = 1000 β 2
Using formula, (x β y)3 = x3 β y3 β 3xy(x β y)
(998)3 = (1000 β 2)3
= (1000)3 β 23 β (3 Γ 1000 Γ 2) (1000 β 2)
= 1000000000 β 8 β 6000(1000 β 2)
= 1000000000 β 8 β 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 β b3 β 12a2b + 6ab2
Solution:
8a3 β b3 β 12a2b + 6ab2 can also be written as (2a)3β b3 β 3(2a)2b + 3(2a)(b)2
8a3 β b3 β 12a2b + 6ab2 = (2a)3 β b3 β 3(2a)2b + 3(2a)(b)2
formula used, (x β y)3 = x3 β y3 β 3xy(x β y)
= (2a β b)3
= (2a β b) (2a β b) (2a β b)
(iii) 27 β 125a3 β 135a + 225a2
Solution:
27 β 125a3 β 135a +225a2 can be also written as 33 β (5a)3 β 3(3)2(5a) + 3(3)(5a)2
27 β 125a3 β 135a + 225a2 = 33 β (5a)3 β 3(3)2(5a) + 3(3)(5a)2
Formula used, (x β y)3 = x3 β y3 β 3xy(x β y)
= (3 β 5a)3
= (3 β 5a) (3 β 5a) (3 β 5a)
(iv) 64a3 β 27b3 β 144a2b + 108ab2
Solution:
64a3 β 27b3 β 144a2b + 108ab2 can also be written as (4a)3 β (3b)3 β 3(4a)2(3b) + 3(4a)(3b)2
64a3 β 27b3 β 144a2b + 108ab2 = (4a)3 β (3b)3 β 3(4a)2(3b) + 3(4a)(3b)2
Formula used, (x β y)3 = x3 β y3 β 3xy(x β y)
= (4a β 3b)3
= (4a β 3b) (4a β 3b) (4a β 3b)
(v) 7p3 β β p2 + p
Solution:
27p3 β β () p2 + ()p can also be written as (3p)3 β β 3(3p)2() + 3(3p)()2
27p3 β () β () p2 + ()p = (3p)3 β ()3 β 3(3p)2() + 3(3p)()2
Formula used, (x β y)3 = x3 β y3 β 3xy(x β y)
= (3p β )3
= (3p β ) (3p β ) (3p β )
Question 9. Verify:
(i) x3 + y3 = (x + y) (x2 β xy + y2)
Solution:
Formula (x + y)3 = x3 + y3 + 3xy(x + y)
x3 + y3 = (x + y)3 β 3xy(x + y)
x3 + y3 = (x + y) [(x + y)2 β 3xy]
x3 + y3 = (x + y) [(x2 + y2 + 2xy) β 3xy]
Therefore, x3 + y3 = (x + y) (x2 + y2 β xy)
(ii) x3 β y3 = (x β y) (x2 + xy + y2)
Solution:
Formula, (x β y)3 = x3 β y3 β 3xy(x β y)
x3 β y3 = (x β y)3 + 3xy(x β y)
x3β y3 = (x β y) [(x β y)2 + 3xy]
x3 β y3 = (x β y) [(x2 + y2 β 2xy) + 3xy]
Therefore, x3 + y3 = (x β 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 β xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z) [(3y)2 β (3y)(5z) + (5z)2]
= (3y + 5z) (9y2 β 15yz + 25z2)
(ii) 64m3 β 343n3
Solution:
64m3 β 343n3 can also be written as (4m)3 β (7n)3
64m3 β 343n3 = (4m)3 β (7n)3
Formula x3 β y3 = (x β y) (x2 + xy + y2)
64m3 β 343n3 = (4m)3 β (7n)3
= (4m β 7n) [(4m)2 + (4m)(7n) + (7n)2]
Question 11. Factorise: 27x3 + y3 + z3 β 9xyz
Solution:
27x3 + y3 + z3 β 9xyz can also be written as (3x)3 + y3 + z3 β 3(3x)(y)(z)
27x3 + y3 + z3 β 9xyz = (3x)3 + y3 + z3 β 3(3x)(y)(z)
Formula, x3 + y3 + z3 β 3xyz = (x + y + z) (x2 + y2 + z2 β xy β yz β zx)
27x3 + y3 + z3 β 9xyz = (3x)3 + y3 + z3 β 3(3x)(y)(z)
= (3x + y + z) [(3x)2 + y2 + z2 β 3xy β yz β 3xz]
= (3x + y + z) (9x2 + y2 + z2 β 3xy β yz β 3xz)
Question 12. Verify that: x3 + y3 + z3 β 3xyz = (1/2) (x + y + z) [(x β y)2 + (y β z)2 + (z β x)2]
Solution:
Formula, x3 + y3 + z3 β 3xyz = (x + y + z)(x2 + y2 + z2 β xy β yz β xz)
Multiplying by 2 and dividing by 2
= (1/2) (x + y + z) [2(x2 + y2 + z2 β xy β yz β xz)]
= (1/2) (x + y + z) (2x2 + 2y2 + 2z2 β 2xy β 2yz β 2xz)
= (1/2) (x + y + z) [(x2 + y2 β 2xy) + (y2 + z2 β 2yz) + (x2 + z2 β 2xz)]
= (1/2) (x + y + z) [(x β y)2 + (y β z)2 + (z β x)2]
Therefore, x3 + y3 + z3 β 3xyz = (1/2) (x + y + z) [(x β y)2 + (y β z)2 + (z β x)2]
Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Solution:
Formula, x3 + y3 + z3 β 3xyz = (x + y + z) (x2 + y2 + z2 β xy β yz β xz)
Given, (x + y + z) = 0,
Then, x3 + y3 + z3 β 3xyz = (0) (x2 + y2 + z2 β xy β yz β xz)
x3 + y3 + z3 β 3xyz = 0
Therefore, x3 + y3 + z3 = 3xyz
Question 14. Without actually calculating the cubes, find the value of each of the following:
(i) (β12)3 + (7)3 + (5)3
Solution:
Let,
x = β12
y = 7
z = 5
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
and we have β12 + 7 + 5 = 0
Therefore, (β12)3 + (7)3 + (5)3 = 3xyz
= 3 Γ -12 Γ 7 Γ 5
= -1260
(ii) (28)3 + (β15)3 + (β13)3
Solution:
Let,
x = 28
y = β15
z = β13
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
and we have, x + y + z = 28 β 15 β 13 = 0
Therefore, (28)3 + (β15)3 + (β13)3 = 3xyz
= 3 (28) (β15) (β13)
= 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 β 35a + 12
Solution:
Using splitting the middle term method,
25a2 β 35a + 12
25a2 β 35a + 12 = 25a2 β 15a β 20a + 12
= 5a(5a β 3) β 4(5a β 3)
= (5a β 4) (5a β 3)
Possible expression for length & breadth is = (5a β 4) & (5a β 3)
(ii) Area : 35y2 + 13y β 12
Solution:
Using the splitting the middle term method,
35y2 + 13y β 12 = 35y2 β 15y + 28y β 12
= 5y(7y β 3) + 4(7y β 3)
= (5y + 4) (7y β 3)
Possible expression for length & breadth is = (5y + 4) & (7y β 3)
Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x2 β 12x
Solution:
3x2 β 12x can also be written as 3x(x β 4)
= (3) (x) (x β 4)
Possible expression for length, breadth & height = 3, x & (x β 4)
(ii) Volume: 12ky2 + 8ky β 20k
Solution:
12ky2 + 8ky β 20k can also be written as 4k (3y2 + 2y β 5)
12ky2 + 8kyβ 20k = 4k(3y2 + 2y β 5)
Using splitting the middle term method.
= 4k (3y2 + 5y β 3y β 5)
= 4k [y(3y + 5) β 1(3y + 5)]
= 4k (3y + 5) (y β 1)
Possible expression for length, breadth & height= 4k, (3y + 5) & (y β 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 Mathsβ 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 β 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 β 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
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