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NCERT Exemplar Class 10 Maths Solutions Chapter 2 Polynomials

NCERT Exemplar Class 10 Maths Solutions Chapter 2 Polynomials

Edited By Ravindra Pindel | Updated on Apr 08, 2025 05:51 PM IST

NCERT exemplar Class 10 Maths Polynomials chapter 2 is the extension of what we have studied in class 9, polynomials. This chapter provides a detailed understanding of Polynomials. A polynomial is an expression that consists of variables, coefficients, with various mathematical operations like addition, subtraction, multiplication, division, and a non-negative integer exponent. The solutions of NCERT exemplar Class 10 Maths chapter 2 important questions are a result of the extensive efforts of subject matter experts and are an excellent source to prepare for NCERT Class 10 Maths. Class 10 Maths chapter 2 mcq are also available in this article.

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  7. NCERT Class 10 Exemplar Solutions - Subject Wise


These NCERT exemplar Class 10 Maths chapter 2 solutions help you to understand the concepts of polynomials with plenty of questions. The CBSE Syllabus Class 10 Maths is the reference for Class 10 Maths NCERT exemplar solutions chapter 2.

Polynomials class 10 MCQ solutions are given below:-

Class 10 Maths chapter 2 exemplar solutions Exercise: 2.1
Page number: 9-10
Total questions: 11

Question:1

If one of the zeroes of the quadratic polynomial (k1)x2+kx+1 is –3, then the value of k is
(A) 43
(B) 43
(C) 23
(D) 23

Answer. [A]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
Let p(x)=(k1)x2+kx+1
If –3 is one of the zeroes of p(x) then p(–3) = 0
put x = –3 in p(x)
p(3)=(k1)(3)2+k(3)+1
0=(k1)(9)3k+1
0=9k93k+1
0=6k8
6k=8
k=86
k=43

Question:2

A quadratic polynomial, whose zeroes are –3 and 4, is
(A) x2x+12
(B)x2+x+12
(C) x22x26
(D) 2x2+2x24

Answer. [C]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is polynomial of degree 2.
(A) p(x)=x2x+12
put x = –3 and put x = 4
p(3)=(3)2(3)+12

=9+3+12=240

p(4)=(4)24+12
=164+12=240
(B) p(x)=x2+x+12
put x = –3 and put x = 4
p(3)=(3)23+12

=9+9=180

p(4)=(4)2+4+12
=16+16=32 0
(C) p(x)=x22x26
put x = –3 and put x = 4
p(3)=(3)22326

=92+326

=9+3122=0

p(4)=(4)22426
=16226
=88=0
(D) p(x)=2x2+2x24
put x = –3 and put x = 4
p(3)=2(3)2+2(3)24

=18624

=120

p(4)=2(4)2+2(4)24
=32+824
=160
If –3, 4 is zeros of a polynomial p(x) then p(–3) = p(4) = 0
Here only (C) option satisfy p(–3) = p(4) = 0

Question:3

If the zeroes of the quadratic polynomial x2+(a+1)x+b are 2 and –3, then
(A) a = –7, b = –1
(B) a = 5, b = –1
(C) a = 2, b = – 6
(D) a = 0, b = – 6

Answer. [D]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a polynomial of degree 2.
If 2 and –3 are the zero of p(x)=x2+(a+1)x+b then p(2) = p(–3) = 0.
p(2)=(2)2+(a+1)(2)+b
0=4+2a+2+b
0=6+2a+b
2a+b=6 ….(1)
p(3)=(3)2+(a+1)(3)+b
0=93a3+b
3ab=6 ….(2)
Add equations (1) and (2)
2a+b+3ab=6+6
2a+3a=0
5a=0
a=0
put a = 0 in (1)
2(0) + b = –6
b=6
Hence a = 0, b = –6.

Question:4

The number of polynomials having zeroes as –2 and 5 is
(A) 1
(B) 2
(C) 3
(D) more than 3

Answer. [D]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Let the polynomial is ax2+bx+c=0 ….(*)
We know that sum of zeroes ba
2+5=ba
31=ba …..(1)
Multiplication of zeroes =ca
2×5=ca
101=ca …..(2)
From equations (1) and (2), it is clear that
a = 1, b = –3, c = –10
Put values of a, b, and c in equation (*)
x33x10=0 ….(3)
But we can multiply or divide equations. (3) by any real number except 0, and the zeroes remain the same.
Hence, there are an infinite number of polynomials that exist with zeroes 2 and 5.
Hence, the answer is more than 3.

Question:5

Given that one of the zeroes of the cubic polynomial ax3+bx2+cx+d is zero, the product of the other two zeroes is
(A) ca
(B) ca
(C) 0
(D) ba

Answer. [B]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a polynomial of degree 3.
Here the given cubic polynomial is ax3+bx2+cx+d
Let three zeroes are α,β,γ
α=0 (given) …..(1)
We know that
αβ+βγ+γα=ca
Put α=0
βγ=ca ( using equation (1))
Hence, the product of the other two zeroes is ca.

Question:6

If one of the zeroes of the cubic polynomial x3+ax2+bx+c is –1, then the product of the other two zeroes is
(A) b – a + 1
(B) b – a – 1
(C) a – b + 1
(D) a – b –1

Answer. [A]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s) and a quadratic polynomial is a polynomial of degree 3.
Here the given cubic polynomial is x3+ax2+bx+c
Let α,β,γ are the zeroes of polynomial
α=1 (given) …..(1)
put x = –1 in p(x)=x3+ax2+bx+c
p(1)=(1)3+a(1)2+b(1)+c
0=1+ab+c(Q=1iszero)
c=1a+b
We know that
αβγ=da
Here, a = 1, b = a, c = b, d = c
So, αβγ=c1
βγ=(1a+b)α
βγ=(1a+b)1=1a+b ( using equation (1))
Hence, the product of the other two is b – a + 1.

Question:7

The zeroes of the quadratic polynomial x2+99x+127 are
(A) both positive
(B) both negative
(C) one positive and one negative
(D) both equal

Answer. [B]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a polynomial of degree 2.
Here, the given quadratic polynomial is x2+99x+127
x2+99x+127
a=1,b=99,c=127
x=b±b24ac2a
x=99±(99)24(1)(127)2(1)
x=99±98015082(1)
x=99±92932
x=99±96.42
x=99+96.42

x=9996.42
x=2.62=1.3

x=97.7
The value of both the zeroes is negative.

Question:8

The zeroes of the quadratic polynomial x2+kx+k,k0,
(A) cannot both be positive
(B) cannot both be negative
(C) are always unequal
(D) are always equal

Answer. [A]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a polynomial of degree 2.
Here, the given quadratic polynomial is x2+kx+k
a=1,b=k,c=k
x=b±b24ac2a
x=k±k24k2
x=k±k(k4)2
k(k4) must be greater than 0
Hence, the value of k is either less than 0 or greater than 4.
If the value of k is less than 0, only one zero is positive.
If the value of k is greater than 4, only one zero is positive.
Hence both the zeroes can not be positive.

Question:9

If the zeroes of the quadratic polynomial ax2+bx+c,c0 are equal, then
(A) c and a have opposite signs
(B) c and b have opposite signs
(C) c and a have the same sign
(D) c and b have the same sign

Answer. [C]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s), and a quadratic polynomial is a polynomial of degree 2.
Here the given polynomial is ax2+bx+c,c0
a=a,b=b,c=c
If both the zeroes are equal, then
b24ac=0
b2=4ac ….(1)
(A) c and a have opposite signs
If c and a have opposite sign, then R.H.S. of equation (1) is negative, but L.H.S. is always positive. So (A) is not a correct one.
(B) c and b have opposite sign
If c is negative and b is positive L.H.S. is positive but R.H.S. of eqn. (1) is negative. Hence, (B) is not the correct one.
(C) c and a have the same sign
If c and a have same sign R.H.S. of eqn. (1) is positive and L.H.S. is always positive hence it is a correct one.
(D) c and b have same sign
If c and b both have negative sign then R.H.S. of eqn. (1) is negative and L.H.S. is positive. So this is not correct.
Only one option i.e. (c) is correct one.

Question:10

If one of the zeroes of a quadratic polynomial of the form x2+ax+b is the negative of the other, then it
(A) has no linear term, and the constant term is negative.
(B) has no linear term and the constant term is positive.
(C) can have a linear term but the constant term is negative.
(D) can have a linear term, but the constant term is positive.

Answer. [A]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
Here the given quadratic polynomial is x2+ax+b …..(1)
a = 1, b = a, c = b
Let x1, x2 are the zeroes of the equation (1)
According to the question:
x2=x1
sum of zeroes = x2+x1=ba
x1x1=ba ( because x2 = - x1 )
0=a1a=0 ( because b = a , a = 1)
Product of zeroes =(x1)(x2)=ca
(x1)(x1)=ca
x12=b ( because c= b, a = 1)
Put the value of a and b in (1)
x2+(x12)
Hence, it has no linear term, and the constant term is negative.

Question:11

Which of the following is not the graph of a quadratic polynomial?
75

Answer:

Answer. [D]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s) and a quadratic polynomial is polynomial of degree 2.
If p(x) is a quadratic polynomial, then there are almost 2 values of x exists called roots.
Hence, the graph of p(x) cuts the x-axis almost 2 times.
But here we see that in option (D) the curve cuts the x-axis at 3 times
Hence, graph (D) is not a graph of a quadratic polynomial.

Class 10 Maths chapter 2 exemplar solutions Exercise: 2.2
Page number: 11-12
Total questions: 2

Question:1

(i)Answer the following and justify:
Can x21 be the quotient on division of x6+2x3+x1 by a polynomial in x of degree 5?
(ii)Answer the following and justify:
What will the quotient and remainder be on division of ax2+bx+c by px3+qx2+rx+s,p0?ax2+bx+c by px3+qx2+rx+s,p0?
(iii)Answer the following and justify:
If on division of a polynomial p (x) by a polynomial g (x), the quotients zero, what is the relation between the degrees of p (x) and g (x)?
(iv) Answer the following and justify:
If on division of a non-zero polynomial p (x) by a polynomial g (x), the remainder is zero, what is the relation between the degrees of p (x) and g (x)?
(v) Answer the following and justify:
Can the quadratic polynomial x2+kx+khave equal zeroes for some odd integer k > 1?

Answer:

(i) Answer. [false]
Solution.
Let divisor of a polynomial in x of degree 5 is = x5+x+1
Quotient = x21
Dividend = x6+2x3+x1
According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x)0, then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by p(x)=g(x)×q(x)+r(x)

i,e. Dividend = Divisor × Quotient + Remainder
=(x5+x+1)(x21)+ Remainder
=x7x5+x3+x+x21+ Remainder
Here it is of degree seven but given dividend is of degree six.
Therefore x21can not be the quotient of x6+2x3+x1 because division algorithm is not satisfied.
Hence, given statement is false.
(ii) Here dividend is ax2+bx+c
and divisor is px3+qx2+rx+s
According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) ¹ 0 then the relation among p(x), g(x) quotient q(x) and remainder r(x) is given by p(x)=g(x)×q(x)+r(x)

where degree of r(x) < degree of g(x).
i.e. Dividend = Devisor × Quotient + Remainder
Here degree of divisor is greater than degree of dividend therefore.
According to division algorithm theorem ax2+bx+c is the remainder and quotient will be zero.
That is remainder = ax2+bx+c
Quotient = 0
(iii) Division algorithm theorem :- According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) is then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by
p(x) = g(x) × q(x) + r(x)
where degree of r(x) < degree of g(x)
i.e. Dividend = Division × Quotient + Remainder
In the given statement it is given that on division of a polynomial p(x) by a polynomial g(x), the quotient is zero.
The given condition is possible only when degree of divisor is greater than degree of dividend
i.e. degree of g(x) > degree of p(x).
(iv)
Division algorithm theorem :- According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) is then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by
p(x) = g(x) × q(x) + r(x) ….(1)
where degree of r(x) < degree of g(x)
According to given statement on division of a non-zero polynomial p(x) by a polynomial g(x) then remainder r(x) is zero then equation (1) becomes
p(x) = g(x) × q(x) ….(2)
From equation (2) we can say g(x) is a factor of p(x) and degree of g(x) may be less than or equal to p(x).

(v)
Answer. [false]
Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)
Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.
Let p(x)=x2+kx+k
It is given that zeroes of p(x) has equal and we know that when any polynomial having equal zeroes than their discriminate is equal to zero
i.e. d=b24ac=0
p(x)=x2+kx+k
here, a = 1, b = k, c = k
d=(k)24×1×k=0
k24k=0
k(k4)=0
k = 0, k = 4
Hence, the given quadratic polynomial x2+kx+k have equal zeroes only when the values of k will be 0 and 4.
Hence, the given statement is not correct.

Question:2

Are the following statements ‘True’ or ‘False’? Justify your answers.
(i)If the zeroes of a quadratic polynomial ax2+bx+c are both positive, then a, b and c all have the same sign
(ii) If the graph of a polynomial intersects the x-axis at only one point, it cannot be a quadratic polynomial.
(iii) If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.
(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.
(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.
(vi) If all three zeroes of a cubic polynomial x3+ax2bx+c are positive, then at least one of a, b and c is non-negative.
(vii) The only value of k for which the quadratic polynomial kx2+x+k has equal zeros is12

(i) Answer. [false]
Polynomial: It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Quadratic polynomial: when the degree of polynomial is two then the polynomial is called quadratic polynomial.
If a quadratic polynomial is p(x)=ax2+bx+c then
Sum of zeroes =ba
Product of zeroes =ca
Here, two possibilities can occurs:
b > 0 and a < 0, c < 0
OR b < 0 and a > 0, c > 0
Here we conclude that a, b and c all have nor same sign is given statement is false.
(ii) Answer. [False]
Polynomial: It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.
We know that the roots of a quadratic polynomial is almost 2, Hence the graph of a quadratic polynomial intersects the x-axis at 2 point, 1 point or 0 point.
For example :
x2+4x+4=0 (a quadratic polynomial)

(x+2)2=0

x+2=0
x=2
Here, only one value of x exist, which is –2.
Hence, the graph of the quadratic polynomial x2+4x+4=0 intersect the x-axis at x = –2.
Hence, we can say that if the graph of a polynomial intersect the x-axis at only one point can be a quadratic polynomial.
Hence, the given statement is false.
(iii)
Answer. [True]
Polynomial: It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.
Let us take an example :-
(x1)2(x2)=0
It is a cubic polynomial
If we find its roots then x = 1, 2
Hence, there are only 2 roots of cubic polynomial (x1)2(x2)=0 exist.
In other words we can say that the graph of this cubic polynomial intersect x-axis at two points x = 1, 2.
Hence we can say that if the graph of a polynomial intersect the x-axis at exactly two points, it need not to be a quadratic polynomial it may be a polynomial of higher degree.
Hence the given statement is true.
(iv)
Answer. [True]
Polynomial: It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Cubic polynomial: when the degree of polynomial is three then the polynomial is called cubic polynomial.
Let α1,α2,α3 be the zeroes of a cubic polynomial.
It is given that two of the given zeroes have value zero.
i.e. α1=α2=0
Let p(y)=(yα1)(yα2)(yα3)
p(y)=(y0)(y0)(yα3)
p(y)=y3y2α3
Here, we conclude that if two zeroes of a cubic polynomial are zero than the polynomial does not have linear and constant terms.
Hence, given statement is true.
(v)
Answer. [True]
Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Cubic polynomial : when the degree of polynomial is three then the polynomial is called cubic polynomial.
Let the standard equation of cubic polynomial is:
p(x)=ax3+bx2+cx+b
Let α,βandγ be the roots of p(x)
It is given that all the zeroes of a cubic polynomial are negative
i.e α=α,β=βandγ=γ
Sum of zeroes =coefficient of x2coefficient of x3
α+β+γ=ba
It is given that zeroes are negative then α=α,β=β, and γ=γ
α+(β)+(γ)=ba
αβγ=ba
(α+β+γ)=ba
(α+β+γ)=ba …….(1)
That is ba>0
Sum of the products of two zeroes at a time =coefficient of xcoefficient of x3
αβ+βγ+γα=ca
Replace α=α,β=βandγ=γ
(α)(β)+(β)(γ)+(γ)(α)=ca
αβ+βγ+γα=ca …..(2)
That is ca>0
Product of all zeroes =-constant termscoefficient of x3
αβγ=da
Replace α=α,β=βandγ=γ
(α)(β)(γ)=da
(αβγ)=da
αβγ=da ……(3)
That is da>0
From equations (1), (2) and (3) we conclude that all the coefficient and the constant term of the polynomial have the same sign.
Hence, the given statement is true.
(vi)
Answer. [False]
Polynomial: It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).
Cubic polynomial: when the degree of polynomial is three then the polynomial is called cubic polynomial.
The given cubic polynomial is
p(x)=x3+ax2bx+c
Let a, b and g are the roots of the given polynomial
Sum of zeroes =coefficient of x2coefficient of x3
α+β+γ=a1=a
We know that when all zeroes of a given polynomial are positive then their sum is also positive
But here a is negative
Sum of the product of two zeroes at a time =coefficient of xcoefficient of x3
αβ+βγ+γα=b1=b
Also, here b is negative
Product of all zeroes =-constant termscoefficient of x3
αβγ=c1=c
Also, c is negative
Hence, if all three zeroes of a cubic polynomial x3+ax2bx+c are positive then a, b and c must be negative.
Hence given statement is false.
(vii)
Answer. [False]
Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
Quadratic polynomial: when the degree of polynomial is two then the polynomial is called quadratic polynomial.
Let p(x)=kx2+x+k
Here it is gives that zeroes of p(x) are equal and we know that when any polynomial having equal zeroes then their discriminate will be equal to zero.
i.e. d = 0
b24ac=0
p(x)=kx2+x+k
Here, a = k, b = 1, c = k
b24ac=0
(1)24×k×k=0
14k2=0
1=4k2
14=k2
k=±12
When +12 and 12 then the given quadratic polynomial has equal zeroes.

Class 10 Maths chapter 2 exemplar solutions Exercise: 2.3
Page number: 12-13
Total questions: 10

Question:1

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
4x23x1

Answer. [1,14]
Zeroes : zeroes of polynomial are the value(s) that makes it equal to 0.
4x23x1=0
4x24x+x1=0
4x(x1)+1(x1)=0
(x1)(4x+1)
x – 1 = 0 or 4x + 1 = 0
x = 1 or 4x = –1
x=14
Hence, 1, 14 are the zeroes of the polynomials
4x23x1
a=4,b=3,c=1
We know that
Sum of the zeroes ba=(3)4=34
Here, 1+(14)=414=34
which is equal to ba
Product of zeroes ca=14
Here 1×14=14=ca

Question:2

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
3x2+4x4

Answer. [2,23]
Solution. Zeroes: zeroes of the polynomial are the value(s) that make it equal to 0.
3x2+4x4=0
3x2+6x2x4=0
3x(x+2)2(x+2)=0
(x+2)(3x2)=0
x+2=0 or 3x2=0
x=2 or x=23
Hence, [2,23] are the zeroes of the polynomial
3x2+4x4
a = 3, b = 4, c = –4
Sum of the zeroes =ba=43
Here, 2+23=6+23=43=ba
Product of the zeroes ca=43
Here, 2×23=43=ca

Question:3

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
5t2+12t+7

Answer. [1,75]
Zeroes : zeroes of the polynomial are the values(s) the makes it equal to 0
5t2+12t+7=0
5t2+5t+7t+7=0
5t(t+1)+7(t+1)=0
(t+1)(5t+7)=0
t + 1 = 0 or 5t + 7 = 0
t = –1 or t=75
Hence, [1,75]are the zeroes of polynomial
Sum of zeroes ba=125
Here, 175=575=125=ba
Product of zeroes ca=75
Here, 1×75=75=ca

Question:4

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
t32t215t

Answer. [0,5,3]
Zeroes : zeroes of the polynomial area the value(s) that makes it equal to 0.
t32t215t =0 (a = 1, b = –2, c = –15, d = 0)
t(t22t15) =0
t(t25t+3t15)=0
t(t(t5)+3(t5))=0
t(t5)(t+3)=0
t = 0 or t – 5 = 0 or t + 3 = 0
t = 5 or t = –3
Hence, 0, 5, –3 are the zeroes of the polynomial
Sum of zeroes =ba=(2)1=2
Here, 0+53=2=ba
Product of zeroes =da=01=0.
Here, 0×5×3=0=da
If α,β,γ are three roots theαβ+βγ+γα=ca
(0×5)+(5×3)+(3×0)=151
15=15
Hence proved.

Question:5

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
2x2+72x+34

Answer. [32,14]
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
2x2+72x+34
Multiply by 4
8x2+14x+3 =0
8x2+12x+2x+3 =0
4x(2x+3)+1(2x+3)=0
(2x+3)(4x+1) =0
2x+3=0 or 4x+1=0
x=32 or x=14
Hence, [32,14] are the zeroes of the polynomial
Here, a=2,b=72,c=34
Sum of zeroes ba=72×2=74
Here, =3214=614=74=ba
Product of zeroes =ca=34×2=38
Here, 32×14=38=ca

Question:6

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
4x2+52x3

Answer. [32,122]
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
4x2+52x3 = 0
4x2+62x2x3=0
22x(2x+3)1(2x+3)=0
(2x+3)(122x)=0
(2x+3)=0 (122x)=0
x=32 or x=122
Hence, [32,122] are the zeroes of the polynomial
Here, a = 4, b = 52 , c = –3
Sum of zeroes =ba=524=522
Here, 32122=6+122=522=ba
Product of zeroes =ca=34
Here, 32×122=34=ca

Question:7

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
2s2(1+22)s+2

Answer. [12,2]
Solution. Zeroes: zeroes of the polynomial are the value(s) that make it equal to 0.
2s2(1+22)s+2 = 0
2s2s22s+2=0
s(2s1)2(2s1)=0
(s2)(2s1)=0

2s1=0 or (s2)=0
s=12 or s=2
Hence, [12,2] are the zeroes of the polynomial
Here, a = 2, b=(1+22),c=2
Sum of zeroes =ba=(1+22)2=1+222
Here, 12+2=1+222
Product of zeroes =ca=22=12
Here, 1×12=12=ca

Question:8

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
v2+43v15

Answer. 53,3
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
v2+43v15=0
v2+53v3v15=0
v(v+53)3(v+53)=0
(v+53)(v3)=0
(v+53)=0 or (v3)=0
v=53 or v=3
Hence, 53,3 are the zeroes of the polynomial
Here, a = 1, b = 43 , c = –15
Sum of zeroes =ba=431=43
Here, 53+3=43=ba
Product of zeroes =ca=15
Here, 53×3=15=ca

Question:9

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
y2+325y5

Answer. [25,52]
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
y2+325y5=0
Multiply by 2
2y2+35y10=0
2y2+45y5y10=0
2y(y+25y)5(y+25y)=0
(y+25y)(2y5)=0
(y+25)=0 or (2y5)=0
y=25 or y=52
Hence, [25,52] are the zeroes of the polynomial
Here, a = 2, b = 35 , c = –10
Sum of zeroes =ba=352
Here, 25+52=45+52=352=ba
Product of zeroes =ca=102=5
Here, 25×52=5=ca

Question:10

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
7y2113y23

Answer. [23,17]
Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.
7y2113y23=0
Multiply by 3
21y211y2=0
21y214y+3y2=0
7y(3y2)+1(3y2)=0
(3y2)(7y+1)=0
3y2=0 or 7y+1=0
y=23 or y=17
Hence, [23,17] are the zeroes of the polynomial
Here, a = 21, b = –11, c = –2
Sum of zeroes ba=(11)21=1121
Here, 2317=14321=1121=ba
Product of zeroes ca=221
Here, 23×17=221=ca

Class 10 Maths chapter 2 solutions Exercise: 2.4
Page number: 14-15
Total questions: 6

Question:1

For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.
(i)83,43
(ii) 218,516
(iii) 23,9
(iv) 325,12

(i) Answer. [2,23]
Solution. Zeroes: zeroes of the polynomial are the value(s) that makes it equal to 0.
Here, sum of zeroes =83
product of zeroes =43
Let p(x) is the required polynomial
p(x) = x2 – (sum of zeroes)x + (product of zeroes)
=x2(83)x+43
=x2+83x+43
Multiply by 3
p(x)=3x2+8x+4
Hence, 3x2+8x+4 is the required polynomial
p(x)=3x2+8x+4
=3x2+6x+2x+4
=3x(x+2)+2(x+2)
=(x+2)(3x+2)=0
x=2,23 are the zeroes of p(x).
(ii) Answer. [52,18]
Solution. Zeroes: zeroes of the polynomial are the value(s) that makes it equal to 0.
Here, sum of zeroes =218
Product of zeroes =516
Let p(x) is the required polynomial
p(x) = x2 – (sum of zeroes)x + (product of zeroes)
=x2(218)x+516
=x2218x+516
Multiply by 16 we get
p(x)=16x242x+5
Hence, 16x242x+5 is the required polynomial
p(x)=16x242x+5
=16x240x2x+5
=8x(2x5)1(2x5)
=(2x5)(8x1)=0
x=52,18 are the zeroes of p(x).
(iii) Answer. [33,3]
Solution. Zeroes: zeroes of the polynomial are the value(s) that makes it equal to 0.
Here, sum of zeroes 23
Product of zeroes = –9
Let p(x) is required polynomial
p(x) = x2 – (sum of zeroes)x + (product of zeroes)
=x2(23)x+(9)
=x2+23x+(9)
p(x)=x2+23x+(9)
Hence, x2+23x+(9) is the required polynomial
p(x)=x2+23x+(9)
=x2+33x3x+(9)
=x(x+33)3(x+33)
=(x+33)(x3)=0
x=33,3 are the zeroes of p(x)
(iv) Answer. [52,15]
Solution. Zeroes: zeroes of the polynomial are the value(s) that makes it equal to 0.
Here sum of zeroes 325
Product of zeroes =12
Let p(x) is the required polynomial
p(x) = x2 – (sum of zeroes)x + (product of zeros)
p(x)=x2(325)x+12
p(x)=x2+325x12
Multiplying by 25 we get
p(x)=25x2+3x5
Hence, 25x2+3x5 is the required polynomial
p(x)=25x2+3x5
=25x2+5x2x5
=5(2x+5)1(2x+5)
=(5x1)(2x+5)
[52,15] are the zeroes of p(x).

Question:2

Given that the zeroes of the cubic polynomial x36x2+3x+10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

Answer. [5, 2, –1]
Solution. Here the given cubic polynomial is x36x2+3x+10
A = 1, B = –6, C = 3, D = 10
Given that, α=a,β=a+b,γ=a+2b
We know that, α+β+γ=BA
a+a+b+a+2b=(6)1
3a+3b=6
3(a + b) = 6
a + b = 2 …..(1)
αβ+βγ+α=CA
a(a+b)+(a+b)(a+2b)+(a+2b)(a)=31
a(2)+2(a+2b)+a2+2ab=3 (Q a + b = 2)
2a+2a+4b+a2+2ab=3
4a+4b+a2+2ab=3
4a+4(2a)+a2+2a(2a)=3 (Q a + b = 2)
4a+84a+a2+4a2a2=3
a2+4a2a2+83=0
a2+4a+5=0
a24a5=0
a25a+a5=0
a(a5)+1(a5)=0
(a5)(a+1)=0
a = 5, –1
Put a = 5 in (1) put a = –1 in (1)
5 + b = 2 –1 + b = 2
b = –3 b = 3
Hence, value of a = 5, b = –3 and a = –1, b = 3
put a = 5, b = –3 and we get zeroes
a = 5
a + b = 5 – 3 = 2
a + 2b = 5 + 2(–3) = –1
Hence, zeroes are 5, 2, –1.

Question:3

Given that 2 is a zero of the cubic polynomial 6x3+2x210x42 , find its other two zeroes.

Answer. [223,12]
Given cubic polynomial is 6x3+2x210x42
If 2 is a zero of the polynomial then (x – 2 ) is a factor of 6x3+2x210x42
76
6x3+2x210x42=(6x2+72x+4)(x2)
=(6x2+42x+32x+42)(x2)
=(2x(3x+22)+2(3x+22))(x2)
=(3x+22)(2x+2)(x2)=0
3x+22=0 2x+2=0
x=223 x=12
Hence, [223,12] are the other two zeroes.

Question:4

Find k so that x2+2x+k is a factor of 2x4+x314x2+5x+6. Also find all the zeroes of the two polynomials.

Answer:

The given polynomial is 2x4+x314x2+5x+6
Here, x2+2x+k is a factor of 2x4+x314x2+5x+6
Use division algorithm
77
Here, x(7k+21)+(6+8k+2k2) is remainder
It is given that x2+2x+k is a factor hence remainder = 0
x(7k+21)+(6+8k+2k2)=0.x+0
By comparing L.H.S. and R.H.S.
7k + 21 = 0
k = – 3 ….(1)
2k2+8k+6=0
2k2+2k+6k+6=0
2k(k + 1) + 6(k + 1) = 0
(k + 1) (2k + 6) = 0
k = –1, –3 …..(2)
From (1) and (2)
k = –3
Hence, 2x4+x314x2+5x+6=(x2+2x+k)(2x23x82k)
Put k = –3
=(x2+2x3)(2x23x8+6)
=(x2+2x3)(2x23x2)
=(x2+3xx3)(2x24x+x2)
=(x(x+3)1(x+3))(2x(x2)+1(x2))
=(x+3)(x1)(x2)(2x+1)
x + 3 = 0 x – 1 = 0 x – 2 = 0 2x + 1 = 0
x = –3 x = 1 x = 2 2x = – 1
x=12
Here zeroes of x2+2x3is –3, 1
Zeroes of 2x23x2is2,12.

Question:5

Given that x5 is a factor of the cubic polynomial x335x2+13x35 ,find all the zeroes of the polynomial.

Answer:

Answer. 5,5+2,52
Solution.
The given cubic polynomial is x335x2+13x35
Hence x5 is a factor
Use divided algorithm
78
x333x2+13x35=(x5)(x225+3)
=(x5)(x225+3)
x225+3
a = 1, b = 25 , c = 3
x=b±b24ac2a
x=25±20122
x=25±222=5±2
x=5+2,52

Hence, 5,5+2,52 are the zeroes of the polynomial.

Question:6

For which values of a and b, are the zeroes of q(x)=x3+2x2+a also the zeroes of the polynomial p(x)=x5x44x3+3x2+3x+b? Which zeroes of p(x) are not the zeroes of q(x)?

Answer: [2, 1]
Solution.
Given :-
q(x)=x3+2x2+a, p(x)=x5x44x3+3x2+3x+b
Hence, q(x) is a factor of p(x) use divided algorithm
79
Since x3+2x2+a is a factor hence remainder = 0
x2(1a)+3x(1+a)+(b2a)=0.x2+0.x+0
By comparing
– 1 – a = 0 b – 2a = 0
a = – 1 2a = b
put a = –1
2(–1) = b
b = –2
Hence, a = –1, b = –2
x5x44x3+3x2+3x2=(x3+2x21)(x23x+2)
=(x3+2x21)(x22xx+2)
=(x3+2x21)(x(x2)1(x2))
=(x3+2x21)(x2)(x1)
x – 2 = 0 x – 1 = 0
x = 2 x = 1
Hence, 2, 1 are the zeroes of p(x), which are not the zeroes of q(x).

NCERT Exemplar Solutions Class 10 Maths Chapter 2 Important Topics:

  • Remainder theorem and zeros of a polynomial.
  • The division method of a polynomial is discussed in detail.
  • The graphical representation of polynomials is discussed in this chapter.
  • NCERT exemplar Class 10 Maths solutions chapter 2 discusses the coefficient of a polynomial and its properties.
  • The relation between zeros of the polynomial and coefficients of the polynomial is also discussed.
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Importance of Solving NCERT Exemplar Class 10 Maths Solutions Chapter 2:

  • These solutions provide a basic knowledge of polynomials, which has great importance in higher classes.

  • The questions based on polynomials can be practiced in a better way, along with these solutions.

  • The NCERT exemplar Class 10 Maths chapter 2 solution polynomials has a good amount of problems for practice and is sufficient for a student.

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