NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 2
Chapter Name : Polynomials

Exercise 2.1

Q1 The graphs of y = p(x) are given in Figure below, for some polynomials p(x). Find the number of
zeroes of p(x), in each case.

Answer. (i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

Page : 28 , Block Name : Exercise 2.1

Exercise 2.2

Q1 Find the zeroes of the following quadratic polynomials and verify the relationship between the
zeroes and the coef cients.
(i) x − 2x − 8
2

(ii) 4s − 4s + 12

(iii) 6x − 3 − 7x 2

(iv) 4u + 8u 2

(v) t − 15 2

(vi) 3x − x − 4 2

Answer. The value of x 2
− 2x − 8 is zero if x +2 = 0 or x —4 0
⇒ x = −2 or x = 4

Therefore, the zeroes of
−(−2) −( Cofficient of x)
2
x − 2x − 8 are— 2and4.N ow( = −2 + 4 = 2 = = 2
1 Cofficient of x
−8 Constant term
= (−2) × 4 = −8 = = 2
1 Cofficient of x

(ii)
2 2
4s − 4s + 1 = (2s − 1)
1
2 s=
The value of 4s − 4s + 1 is zero when 2s − 1 = 0, i.e., 2

2 1 1
Therefore, the zeroes of 4s − 4s + 1 are and
2 2

−(−4) −( Coefficient of s)
Sum of zeroes = 1

2
+
1

2
= 1 =
4
=
( Coefficient of s )
2

Product of zeroes= 1
×
1
=
1
=
Constant term

2 2 4 Coefficient of s2

(iii)
2 2
6x − 3 − 7x = 6x − 7x − 3 = (3x + 1)(2x − 3)
2
The value of 6x − 3 − 7x is zero when 3x + 1 = 0 or 2x − 3 = 0, i.e.,

−1 3
x = or x =
3 2

2 3 3
Therefore, the zeroes of 6x − 3 − 7x are and
3 2

−1 3 7 −(−7) −( Coefficient of x)
Sum of zeroes = + = = =
3 2 6 6 6

−1 3 −1 −3 Constant term
Product of zeroes = × = = =
3 2 2 6 Coefficient of x2

(iv)
2 2
4u + 8u = 4u + 8u + 0

= 4u(u + 2)
2
The value of 4u + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or

u = −2

2
Therefore, the zeroes of 4u + 8u are 0 and − 2 .
−(8)
−( Coefficient of u)
Sum of zeroes= 0+(−2)=−2=
4 =
Coefficient of u2

Product of zeroes= 0×(−2)
= 0 =
0

4
=
Constant term

Coefficient of u
2

(v)

2
t − 15

2
= t − 0.1 − 15

= (t − √15)(t + √15)
2
The value of t − 15 is zero when t − √15 = 0 or t + √15 = 0, l.e., when

t = √15 or t = −√15

2
Therefore, the zeroes of t − 15 are √15 and - √15 .
−( Coefficient of t)
Sum of zeroes =√15 + (−√15) = 0 =
−0
= 2
1 ( Coefficient of t )

Product of zeroes=( √15)(−√15) = −15 =
−15 Constant term
= 2
1 Coefficient of x

2
(vi) 3x − x − 4

= (3x − 4)(x + 1)

2
The value of 3x − x − 4 is zero when 3x − 4 = 0 or x + 1 = 0, i.e.,

4
when x = or x = −1
3
4
2
Therefore, the zeroes of 3x − x − 4 are 3 and − 1

4 1 −(−1) −( Coefficient of x)
Sum of zeroes = + (−1) = = = 2
3 3 3 Coefficient of x

Product of zeroes= 4 −4 Constant term
(−1) = = 2
3 3 Coefficient of x

Page : 33 , Block Name : Exercise 2.2

Q2 Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes
respectively.
(i) , −11

4

(ii) √2, 1

3

(iii) 0, √5
(iv) 1,1
(v) − , 1

4
1

4

(vi) 4, 1

Answer. (i) Let and ß are the zeroes of the polynomial ax 2
+ bx + c , then we have

1 −b
α + β = =
4 a

On comparing, αβ = −1 =
−4 C
=
4 a

a=4,b=-1,and c=-4

(ii) Let and ß are the zeroes of the polynomial ax 2
+ bx + c , then we have
3√2 −b
α + β = √2 = =
3 a

On comparing,
1 c
αβ = =
3 a

a = 3, b = −3√2 and c = 1

Hence, the required quadratic polynomial is x 2
+ 0. x + √5

(iii) Let a and are the zeroes or the polynomial ax 2
+ bx + c . then we have

0 −b
α + β = 0 = =
1 a

√5
on comparing αβ = √5 =
c
=
1 a

Hence, the required quadratic polynomial is : x 2
+ 0. x + √5

(iv) Let a and are the zeroes Of the polynomial ax 2
+ bx + c , then we have
1 −b
α + β = 1 = =
1 a
1 c
αβ = 1 = =
1 a

on comparing, a = 1, b = −1 and c = 1
Hence. the required quadratic polynomial is x 2
− x + 1

(v) Let a and are the zeroes Of the polynomial ax 2
+ bx + c , then we have
−1 −b
α + β = 1 = =
4 a

1 c
αβ = 1 = =
4 a

on comparing,
a=4,b=1,and c=1
Hence. the required quadratic polynomial is 4x 2
+ x + 1

(vi) Let a and are the zeroes Ofthe polynomial axa + bx + c, then we have
4 −b
α + β = 4 = =
1 a

1 c
αβ = 1 = =
1 a

on comparing ,
a=1, b=-4 and c=1
Hence, the required quadratic polynomial is x 2
− 4x + 1

Page : 33 , Block Name : Exercise 2.2

Exercise 2.3

Q1 Divide the polynomial p(x) by the polynomial g(x) and nd the quotient and remainder in each
of the following :

3 2 2
p(x) = x − 3x + 5x − 3, g(x) = x − 2
4 2 2
p(x) = x − 3x + 4x + 5, g(x) = x + 1 − x
4 2
p(x) = x − 5x + 6, g(x) = 2 − x

Answer. (i) p(x) = x 3
− 3x
2
+ 5x − 3, g(x) = x
2
− 2

(ii)
4 2 4 3 2
p(x) = x − 3x + 4x + 5 = x + 0. x − 3x + 4x + 5
2 2
q(x) = x + 1 − x = x − x + 1

(iii)
4 4 2
p(x) = x − 5x + 6 = x + 0x − 5x + 6
2 2
q(x) = 2 − x = −x + 2

Page : 36 , Block Name : Exercise 2.3

Q2 Check whether the rst polynomial is a factor of the second polynomial by dividing the second
polynomial by the rst polynomial:
2 4 3 2
(i)t − 3, 2t + 3t − 2t − 9t − 12

(ii) x + 3x + 1, 3x + 5x − 7x + 2x + 2
2 4 3 2

(iii) x − 3x + 1, x − 4x + x + 3x + 1
3 5 3 2

Answer. (i)t 2
− 3, 2t
4
+ 3t
3
− 2t
2
− 9t − 12
2 2
t − 3 = t + 0. t − 3

Since the remainder is o,
Hence,t − 3 is a factor of 2t
2 4
+ 3t
3
− 2t
2
− 9t − 12

(ii)
2 4 3 2
x + 3x + 1, 3x + 5x − 7x + 2x + 2

(iii)
3 5 3 2
x − 3x + 1, x − 4x + x + 3x + 1

Page : 36 , Block Name : Exercise 2.3

Q3 Obtain all other zeroes of 3x 4
+ 6x
3
− 2x
2
− 10x − 5 , if two of its zeroes are √ 5

3
and − √
5

3
.

Answer. p(x) = 3x 4
+ 6x
3
− 2x
2
− 10x − 5

Since the two zeroes are √ 5

3
and − √
5

3
,

∴ (x − √
5

3
) (x + √
5

3
) = (x
2
−
5

3
) is a factor of 3x 4
+ 6x
3
− 2x
2
− 10x − 5 .

Therefore, we divide the given polynimial by x .
2 5
−
3

Hence, the zeroes of the given polynomial are √ 5

3
, −√
5

3
, −1 and − 1 .

Page : 36 , Block Name : Exercise 2.3

Q4 On dividing x − 3x + x + 2 by a polynomial g(x), the quotient and remainder were x-2 and -2
3 2

x+4, respectively. Find g(x).

Answer.
3 2
p(x) = x − 3x + x + 2 (Dividend)

g(x) =? (Divisor)

Quotient = (x − 2)

Remainder = (−2x + 4)

Dividend = Divisor * Quotient + Remainder

3 2
x − 3x + x + 2 = g(x) × (x − 2) + (−2x + 4)

3 2
x − 3x + x + 2 + 2x − 4 = g(x)(x − 2)

3 2
x − 3x + 3x − 2 = g(x)(x − 2)

3 2
g(x) is the quotient when we divide (x − 3x + 3x − 2) by (x − 2)

Page : 36 , Block Name : Exercise 2.3

Q5 Give examples of polynomial p(x),g(x),q(x) and r(x) which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

Answer.
According to the division algorithm, if p(x) and g(x) are two

polynomials with

g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that

p(x) = g(x) × q(x) + r(x)

where r(x) = 0 or degree of r(x) < degree of the variable in the

polynomial.

(i) deg p(x) = deg q(x)

Degree of quotient will be equal to degree of dividend when divisor is

constant (i.e., when any polynomial is divided by a constant).

2
Let us assume the division of 6x + 2x + 2 by 2 .

2
Here, p(x) = 6x + 2x + 2

g(x) = 2

2
q(x) = 3x + x + 1 and r(x) = 0

Degree of p(x) and q(x) is the same i.e., 2 .

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)
2 2
6x + 2x + 2 = 2 (3x + x + 1)

2
= 6x + 2x + 2

Thus, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)

3 2
Let us assume the division of x + x by x ,

3
Here, p(x) = x + x

2
g(x) = x

q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

3 2
x + x = (x ) × x + x

3 3
x + x = x + x

Thus, the division algorithm is satisfied.

(iii) deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.

3 2
Let us assume the division of x + 1 by x .

3
Here, p(x) = x + 1

2
g(x) = x

q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0 .

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

3 2
x + 1 = (x ) × x + 1

3 3
x + 1 = x + 1

Thus, the division algorithm is satisfied.

Page : 36 , Block Name : Exercise 2.3

Exercise 2.4

Q1 Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also
verify the relationship between the zeroes and the coef cients in each case:
(i) 2x + x − 5x + 2; , 1, −2
3 2 1

2

(ii) x 3
− 4x
2
+ 5x − 2; 2, 1, 1

Answer.
3 2
(i) P (x) = 2x + x − 5x + 2

1
Zeroes for this polynomial are , 1, −2
2

3 2
1 1 1 1
P ( ) = 2( ) + ( ) − 5( ) + 2
2 2 2 2

1 1 5
= + − + 2
4 4 2

= 0

3 2
p(1) = 2 × 1 + 1 − 5 × 1 + 2

= 0

3 2
p(−2) = 2(−2) + (−2) − 5(−2) + 2

= −16 + 4 + 10 + 2 = 0

1
Therefore, , 1, and − 2 are the zeroes of the given polynomial.
2

3 2
Comparing the given polynomial with ax + bx + cx + d, we obtain a = 2,

b = 1, c = −5, d = 2

1
We can take α = , β = 1, y = −2
2

1 1 −b
α + β + γ = + 1 + (−2) = − =
2 2 a

1 1 −5 c
αβ + βγ + αγ = × 1 + 1(−2) + (−2) = =
2 2 2 a

−1 −(2) −d
1
αβγ = × 1 × (−2) = = =
2 1 2 a

Therefore, the relationship between the zeroes and the coefficients is

verified.

3 2
(ii) p(x) = x − 4x + 5x − 2

Zeroes for this polynomial are 2, 1, 1 .

3 2
p(2) = 2 − 4 (2 ) + 5(2) − 2

= 8 − 16 + 10 − 2 = 0

3 2
p(1) = 1 − 4(1) + 5(1) − 2

= 1 − 4 + 5 − 2 = 0

Therefore, 2, 1, 1 are the zeroes of the given polynomial.

3 2
Comparing the given polynomial with ax + bx + cx + d, we obtain a = 1

b = −4, c = 5, d = −2

Verification of the relationship between zeroes and coefficient of the

given polynomial

−(−4) −b
Sum of zeroes 2 + 1 + 1 = 4 = =
1 a

Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1)
(5) c
= =
= 2 + 1 + 2 = 5 1 a

−(−2) −d
= =
Multiplication of zeroes = 2 × 1 × 1 = 2 1 a

Hence, the relationship between the zeroes and the coefficients is

verified.

Page : 36 , Block Name :Exercise 2.4

Q2 Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time. and
the product of its zeroes as 2, -7, -14 respectively.

Answer.
3 2
Let the polynomial be ax + bx + cx + d and the zeroes be α, β, and γ

It is given that

2 −b
α + β + γ = =
1 a

−7 c
αβ + βγ + αγ = =
a a

−14 −d
αβγ = =
1 a

If a = 1, then b = −2, c = −7, d = 14

3 2
Hence, the polynomial is x − 2x − 7x + 14.

Page : 36 , Block Name : Exercise 2.4

Q3 If the zeroes of the polynomial x 3
− 3x
2
+ x + 1 are a − b, a, a + b, find a and b.

Answer.
3 2
p(x) = x − 3x + x + 1

Zeroes are a-b,a+a+b
comparing the given polynomial with px 3
+ qx
2
+ rx + t ,we obtain
p=1,q=-3,r=1,t=1
sum of zeroes = a-b+a+a+b
−q
= 3a
p

−(−3)
= 3a
1

3=3a
a=1
The zeroes are 1-b,1,1+b
Multiplication of zeroes =1(1-b)(1+b)
−t 2
= 1 − b
p

−1 2
= 1 − b
1
2
1 − b = −1

2
1 + 1 = b

b = ± √2

Page : 37 , Block Name : Exercise 2.4

Q4 If two zeroes of the polynomial x 4
− 6x
3
− 26x
2
+ 138x − 35 are 2 ± √3, nd other zeroes.

Answer. Given that 2 + √3 and 2 − √3 are zeroes of the given polynomial.
Therefore, (x − 2 − √3)(x − 2 + √3) = x + 4 − 4x − 3 2

=x − 4x + 1 is a factor of the given polynomial
2

For nding the remaining zeroes of the given polynomial, we will nd the quotient by dividing
4 3 2 2
x − 6x − 26x + 138x − 35 by x − 4x + 1

Page : 37 , Block Name : Exercise 2.4

Q5
4 3 2 2
If the polynomial x − 6x + 16x − 25x + 10 is divided by another polynomial x − 2x + k

the remainder comes out to be x + a, find k and a.

Answer.
By division algorithm,

Dividend = Divisor × Quotient + Remainder

Dividend - Remainder = Divisor × Quotient

4 3 2 4 3 2
x − 6x + 16x − 25x + 10 − x − a = x − 6x + 16x − 26x + 10 − a will be perfectly

2
divisible by x − 2x + k .

4 3 2 2
Let us divide x − 6x + 16x − 26x + 10 − abyx − 2x + k

Page : 37 , Block Name : Exercise 2.4