 Lesson: Polynomials
EXERCISE 2.1 (Multiple Choice Questions and
Choose the correct answer from the given four
options in the following questions:
Question: 1
If one of the zeroes of the quadratic polynomial
is , then the value of is:
2
(k 1)x kx 1
3
k
(A)
4
3
(B)
4
3
(C)
2
3
(D)
2
3
Solution
(A)
Let
2
p x (k 1)x kx 1
 
2
(k 1) 3 k 3 1 0
9 k 1 3k 1 0
9k 9 3k 1 0 6k 8 0
8
k
6
4
k
3
Question: 2
A quadratic polynomial, whose zeroes are and ,
3
4
is:
(A)
2
x x 12
(B)
2
x x 12
(C)
2
x x
6
2 2
(D)
2
2x 2x 24
Solution
(C)
Sum of zeroes
3 4
1
1
1
b
a
Product of zeroes
3 4
12
12
1 c
a
Now, and
a 1,
b 1
c 12
Required polynomial
2
ax bx c
2
1 x 1 x 12
2
x x
6
2 2
Question: 3
If the zeroes of the quadratic polynomial
are and , then
2
x a 1 x b
2
3
(A)
a 7,b 1
(B)
a 5,b 1
(C)
a 2,b 6
(D)
a 0,b 6
Solution
(D)
Let
2
p x x a 1 x b
p 2 0
2
2 a 1 2 b 0
4 2a 2 b 0 2a b 6
2
3 a 1 3 b 0
9 3a 3 b 0
3a b 6
Now, solve equations and .
2a b 6
3a b 6
5a 0
a 0
2 0 b 6
b 6
Thus, and .
a 0
b 6
Question: 4
The number of polynomials having zeroes as and
2
is
5
(A)
1
(B)
2
(C)
3
(D) more than
3
Solution
(D) Sum of zeroes
2 5
3
1
3
1
b
a
Product of zeroes
2 5
10
1
c
a
Now,
a 1,b 3, c 10
Required polynomial
2
ax bx c
2
1 x 3 x 10
2
x 3x 10
When we multiply, or divide any polynomial by any
arbitrary constant, the zeroes of the polynomial
remain the same.
Multiply by ,
k
2
p x kx 3kx 10k
where, is a real number.
k
Divide by ,
k 2
x 3x 10
p x
k k k
where, is a non-zero real number.
k
Thus, the required number of polynomials is infinite
i.e. more than 3.
Question: 5
If one of the zeroes of the cubic polynomial
is zero, the product of then other
3 2
ax bx cx d
two zeroes is:
(A)
c
a
(B)
c
a
(C)
0
(D)
b
a
Solution
(B)
Let
3 2
p x ax bx cx d
Let , and be the zeroes of cubic polynomial
, where .
p x
0
Sum of the product of two zeroes taken at a time
c
a 
c
a
c
0 0
a
c
a
Thus, the product of the other two zeroes is .
c
a
Question: 6
If one of the zeroes of the cubic polynomial
is , then the product of the other
3 2
x ax bx c
1
two zeroes is
(A)
b a 1
(B)
b a 1
(C)
a b 1
(D)
a b 1
Solution
(A)
Let
3 2
p x x ax bx c
Let , and be the zeroes of cubic polynomial
, where .
p x
1
p 1 0
3 2
1 a 1 b 1 c 0
1 a b c 0
c 1 a b We know,
3
Constant term
Product of zeroes
Coefficient of x
c
1
 c
1 c
c
1 a b
Thus, the product of the other two zeroes is .
1 a b
Question: 7
The zeroes of the quadratic polynomial
2
x 99x 127
are:
(A) both positive
(B) both negative
(C) one positive and one negative
(D) both equal
Solution
(B)
Let .
2
p x x 99x 127
Compare with .
p x
2
ax bx c a 1,b 99,c 127
Let and be the zeroes of the polynomial .
p x
Sum of zeroes,
b
a
99
1
99
Product of zeroes,

c
a
127
1
127
When the product of zeroes is positive, then either
both the zeroes are negative or positive. When the
sum of these zeroes is negative, then the zeroes must
be negative.
Thus, both zeroes of the given polynomial are
negative.
Question: 8
The zeroes of the quadratic polynomial
2
, 0 x kx k k (A) cannot both be positive
(B) cannot both be negative
(C) are always unequal
(D) are always equal
Solution
(A)
Let
2
p x x kx k,k 0
Compare with .
p x
2
ax bx c
a 1,b k,c k
Let and be the zeroes of the polynomial .
p x
We know,
Sum of zeroes,
1
k
k
Product of zeroes
1

k
k
There are two cases.
Case 1: is negative
k
When is negative, is also negative.
k

It means and are of opposite signs.
Case 2: is positive
k When is positive, is also positive, but is
k

negative.
When the product of zeroes is positive, then either
both the zeroes are negative or both are positive.
When the sum of these zeroes is negative, then the
zeroes must be negative.
Thus, in any case the zeroes of the given polynomial
can’t both be positive.
Question: 9
If the zeroes of the quadratic polynomial
are equal, then
2
ax bx c, c 0
(A) and have opposite signs
c
a
(B) and have opposite signs
c
b
(C) and have the same sign
c
a
(D) and have the same sign
c
b
Solution
(C)
Let
2
p x ax bx c
Let and be the zeroes of the polynomial .
p x
When , then they have the same sign (both are
positive or both are negative). For ,
 0
Product of zeroes,

c
a
So,
c
0
a
This is possible only when and both have the
a
c
same sign.
Question: 10
If one of the zeroes of a quadratic polynomial of the
form is the negative of the other, then it
2
x ax b
(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.
Solution
(A)
Let
2
p x x ax b
Let and be the zeroes of the given polynomial.
Sum of the zeroes 0
1
a
0a
Now, , that cannot be linear.
2
p x x b
Product of the zeroes
b
2
b
This is possible when .
b 0
Thus, it has no linear term and the constant term is
negative.
Question: 11
Which of the following is not the graph of a Solution
(D)
For any quadratic polynomial, the graph has one of
the two shapes; it either opens upwards or opens
downwards , depending upon whether or
a 0
. These curves are known as parabolas.
a 0
Thus, option (D) cannot be possible.
EXERCISE 2.2
Question: 1 (i) Can be the quotient on division of
2
x 1
by a polynomial in of degree ?
6 3
x 2x x 1
x
5
(ii) What will the quotient and remainder be on
division of by
2
ax bx c
3 2
px qx rx s, p 0
?
(iii) If on division of a polynomial by a
p x
polynomial , the quotient is zero, what is
g x
the relation between the degrees of and
p x
?
g x
(iv) If on division of a non-zero polynomial by
p x
a polynomial , the remainder is zero, what
g x
is the relation between the degrees of and
p x
?
g x
(v) Can the quadratic polynomial have
2
x kx k
equal zeroes for some odd integer ?
k 1
Solution
(i) No.
When you divide the polynomial
6 3
x 2x x 1
by a polynomial of degree 2, then the degree of
the polynomial of quotient is 4.
By the division algorithm,
Dividend Divisor Quotient Remainder
Degree of divisor + Degree of quotient = Degree
of dividend
(ii) Divisor
3 2
px qx rx s,p 0
Dividend
2
ax bx c
When the degree of dividend the degree of
divisor, then the quotient will be zero and
remainder is the same as the dividend.
(iii) The relation between the degrees of and
p x
is:
g x
The degree of is less than the degree of
p x
.
g x
(iv) Thus is a factor of .
g x
p x
Also, the degree of is less than or equal to
g x
the degree of .
p x
(v) No.
Let
2
p x x kx k
Let and be the zeroes of the polynomial
.
p x
We know, Sum of zeroes
b
a
2
1
k
k
…………………..(1)
2
k
Product of zeroes
c
a
2
1
k
………………………………(2)
k
Now, substitute the value of from equation (1)
into (2).
2
k
k
4
2
k 4k
2
k 4k 0
k k 4 0
or
0k
4k
But .
k 1
Thus, ; this is an even, not an odd number.
k 4 Question: 2
Are the following statements ‘True’ or ‘False’? Justify
(i) If the zeroes of a quadratic polynomial
are both positive, then and all
2
ax bx c
a,b
c
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
are positive, then at least one of
3 2
x ax bx c
and is non-negative.
a,b
c (vii) The only value of for which the quadratic
k
polynomial has equal zeros is .
2
kx x k
1
2
Solution
(i) False.
The zeroes of a quadratic polynomial
are both positive then
2
ax bx c
and
b
a

c
a
where are zeroes of the quadratic
and
polynomial.
If and then
c 0
a 0
b 0
Or and then
c 0
a 0
b 0
(ii) False.
A quadratic polynomial may touch the -axis at
x
exactly one point or intersects -axis at exactly
x
two points or does not touch the -axis.
x
(iii) True.
If the graph of a polynomial intersects the -
x
axis at exactly two points, then it may or may
not be a quadratic polynomial because a
polynomial of degree more than 2 is possible
that intersects the -axis at exactly two points.
x (iv) True.
Let , and be the zeroes of the cubic
polynomial.
Let
0
Also,
p x x x x
x x 0 x 0
3 2
x x
This does not have linear and constant terms.
(v) True.
Let , and be the zeroes.
b
a
Here, and have the same sign.
a
b

d
a
Here, have the same sign.
a,d
 
c
a
Here, have the same sign.
a,c
So, have the same sign.
a,b,c,d
(vi) False.
Let , and be the three zeroes of the
polynomial .
3 2
x ax bx c Product of zeroes

c
1
 c
All three zeroes are positive. So, the product of
all the three zeroes should also be positive.
 0
c 0
c 0
Sum of zeroes
a
1
a
However, , and are all positive.
0
a 0
a 0
Sum of the product of two zeroes at a time
b
1

b
1
 0
b
0
1
b 0 Thus, the cubic polynomial has three positive
zeroes when all constants , and are
a
b
c
negative.
(vii) False
Let and be the two zeroes.
Sum of zeroes
b
a
1
2
k
……………..(1)
1
2
k
Product of zeroes
c
a
2
k
k
……………………(2)
1
Solve the equation (1) and (2).
2
1
1
4k
2
4k 1
2
1
k
4
1
k
2 Now, or
1
k
2
1
k
2
Thus, for two values of , the given polynomial
k
has equal zeroes.
EXERCISE 2.3
Find the zeroes of the following polynomials by
factorisation method and verify the relations
between the zeroes and the coefficients of the
polynomials
Question: 1
2
4x 3x 1
Solution
Let
2
p x 4x 3x 1
2 2
4x 3x 1 4x 4x x 1
4x x 1 1 x 1
x 1 4x 1
Now,
or
x 1 0
4x 1 0
or
x 1
1
x
4 So, the zeroes are and .
1
1
4
Sum of zeroes
1
1
4
3
4
3
4
2
coefficient of x
coefficient of x
Product of zeroes
1
1
4
1
4
2
constant term
coefficient of x
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
Question: 2
2
3x 4x 4
Solution:
Let
2
p x 3x 4x 4
2 2
3x 4x 4 3x 6x 2x 4 3x x 2 2 x 2
x 2 3x 2
Now,
or
x 2 0
3x 2 0
or
x 2
2
x
3
So, the zeroes are and .
2
2
3
Sum of zeroes
2
2
3
4
3
2
coefficient of x
coefficient of x
Product of zeroes
2
2
3
4
3
2
constant term
coefficient of x
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified. Question: 3
2
5t 12t 7
Solution:
Let
2
p t 5t 12t 7
2 2
5t 12t 7 5t 5t 7t 7
5t t 1 7 t 1
t 1 5t 7
Now,
or
t 1 0
5t 7 0
or
t 1
7
t
5
So, the zeroes are and .
1
7
5
Sum of zeroes
7
1
5
12
5
2
coefficient of t
coefficient of t
Product of zeroes 7
1
5
7
5
2
constant term
coefficient of t
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
Question: 4
3 2
t 2t 15t
Solution:
Let
3 2
p t t 2t 15t
3 2 2
t 2t 15t t t 2t 15
2
t t 5t 3t 15
t t t 5 3 t 5
t t 5 t 3
Now,
or or
t 0
t 5 0
t 3 0
or or
t 0
t 5
t 3
So, the zeroes are and .
0,5
3
Sum of zeroes 0 5 3
2
2
1
2
3
coefficient of t
coefficient of t
Sum of product of zeroes taking two at a time
0 5 5 3 3 0  
15
0 15 0
1
3
coefficient of t
coefficient of t
Product of zeroes
0 5 3
0 0
0
1 1
3
constant term
coefficient of t
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
Question: 5
2
7 3
2x x
2 4 Solution
Let
2
7 3
p x 2x x
2 4
2 2
7 3 1
2x x 8x 14x 3
2 4 4
2
1
8x 12x 2x 3
4
1
4x 2x 3 1 2x 3
4
1
2x 3 4x 1
4
Now,
or
2x 3 0
4x 1 0
or
3
x
2
1
x
4
So, the zeroes are and .
3
2
1
4
Sum of zeroes
3 1
2 4
7
4
2
coefficient of x
coefficient of x
Product of zeroes 3 1
2 4
3
8
2
constant term
coefficient of x
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
Question: 6
2
4x 5 2x 3
Solution:
Let
2
p x 4x 5 2x 3
2 2
4x 5 2x 3 4x 6 2x 2x 3
2 2 2 3 1 2 3 x x x
2x 3 2 2x 1
Now,
or
2x 3 0
2 2x 1 0
or
3
x
2
1
x
2 2 So, the zeroes are and .
3
2
1
2 2
Sum of zeroes
3 1
2 2 2
6 1
2 2
5
2 2
5 2
4
2
coefficient of x
coefficient of x
Product of zeroes
3 1
2 2 2
3
4
2
constant term
coefficient of x
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
Question: 7 2
2s 1 2 2 s 2
Solution:
Let
2
p s 2s 1 2 2 s 2
2 2
2s 1 2 2 s 2 2s s 2 2s 2
  s 2s 1 2 2s 1
2s 1 s 2
Now,
or
2s 1 0
s 2 0
or
1
s
2
s 2
So, the zeroes are and .
1
2
2
Sum of zeroes
1
2
2
1 2 2
2
1 2 2
2
2
coefficient of s
coefficient of s Product of zeroes
1
2
2
1
2
2
constant term
coefficient of s
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
Question: 8
2
v 4 3v 15
Solution:
Let
2
p v v 4 3v 15
2 2
v 4 3v 15 v 5 3v 3v 15
v v 5 3 3 v 5 3
v 5 3 v 3
Now,
or
v 5 3 0
v 3 0
or
v 5 3
v 3 So, the zeroes are and .
5 3
3
Sum of zeroes
5 3 3
4 3
4 3
1
2
coefficient of v
coefficient of v
Product of zeroes
5 3 3
5 3
15
15
1
2
constant term
coefficient of v
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
Question: 9
2
3
y 5y 5
2 Solution:
Let
2
3
p y y 5y 5
2
2 2
3 1
5 5 2 3 5 10
2 2

y y y y
2
1
2y 4 5y 5y 10
2
1
2y y 2 5 5 y 2 5
2
1
2y 5 y 2 5
2
Now,
or
2y 5 0
y 2 5 0
or
5
y
2
y 2 5
So, the zeroes are and .
2 5
5
2
Sum of zeroes
5
2 5
2
3 5
2
3
5
2 2
coefficient of y
coefficient of y
Product of zeroes
5
2 5
2
5
5
1
2
constant term
coefficient of y
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
Question: 10
2
11 2
7y y
3 3
Solution
Let
2
11 2
p y 7y y
3 3
 
2 2
11 2 1
7y y 21y 11y 2
3 3 3
2
1
21y 14y 3y 2
3 1
7y 3y 2 1 3y 2
3
1
3y 2 7y 1
3
Now,
or
3y 2 0
7y 1 0
or
2
y
3
1
y
7
So, the zeroes are and .
2
3
1
7
Sum of zeroes
2 1
3 7
11
21
11
3 7
2
coefficient of y
coefficient of y
Product of zeroes
2 1
3 7
2
21 2
3 7
2
3
7
2
constant term
coefficient of y
Thus, the relationship between zeroes and the
coefficients of the polynomial is verified.
EXERCISE 2.4
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)
8 4
,
3 3
(ii)
21 5
,
8 16
(iii)
2 3, 9
(iv)
3 1
,
2
2 5
Solution: (i)
8 4
,
3 3
Sum of zeroes,
8
3
Product of zeroes,
4
3

Required equation

2
p x x
2
8 4
x x
3 3
or
2
1
3x 8x 4
3
2
3x 8x 4
Factorisation by splitting method,
2 2
3x 8x 4 3x 6x 2x 4
3x x 2 2 x 2
x 2 3x 2
Thus, the zeroes are and .
2
2
3
(ii)
21 5
,
8 16
Sum of zeroes,
21
8
Product of zeroes,
5
16

Required equation

2
p x x 2
21 5
x x
8 16
or
2
1
16 42 5
16
x x
2
16 42 5 x x
Factorisation by splitting method,
2 2
16x 42x 5 16x 40x 2x 5
8x 2x 5 1 2x 5
2x 5 8x 1
Thus, the zeroes are and .
5
2
1
8
(iii)
2 3, 9
Sum of zeroes,
2 3
Product of zeroes.
9
Required equation

2
p x x
2
x 2 3 x 9
2
x 2 3x 9
Factorisation by splitting method,
2 2
x 2 3x 9 x 3 3x 3x 9
x x 3 3 3 x 3 3
x 3 3 x 3
Thus, the zeroes are and .
3 3
3 (iv)
3 1
,
2
2 5
Sum of zeroes,
3
2 5
Product of zeroes,
1
2

Required equation

2
p x x
2
3 1
x x
2
2 5
or
2
1
2 5x 3x 5
2 5
2
2 5x 3x 5
Factorisation by splitting method,
2 2
2 5x 3x 5 2 5x 5x 2x 5
5x 2x 5 1 2x 5
2x 5 5x 1
Thus, the zeroes are and .
5
2
1
5
Question: 2
Given that the zeroes of the cubic polynomial
are of the form for
3 2
x 6x 3x 10
a,a b,a 2b
some real numbers and , find the values of and
a
b
a
as well as the zeroes of the given polynomial.
b Solution:
Let
3 2
p x x 6x 3x 10
Let , and are the zeroes of .
a
a b
a 2b
p x
2
3
coefficient of x
Sum of thezeroes
coefficient of x
6
a a b a 2b
1
3a 3b 6
a b 2
3
Sum of theproduct of
coefficient of x
zeroes taking twoat a time
coefficient of x
3
a a b a b a 2b a a 2b
1
3
a a b a b a b b a a b b
1
Substitute the value of ,
a b
a a b a b a b b a a b b 3
2a 2 2 b a 2 b 3
2a 2 2 2 a a 2 2 a 3
2
2a 8 2a 4a a 3
2
a 8 3 4a 0
2
a 4a 5 0
Factorisation by splitting method,
2
a 4a 5 0
2
a 5a a 5 0
a a 5 1 a 5 0
a 5 a 1 0
So, or
a 1
5
When , then
a 1
b 3
When , then
a 5
b 3
The zeroes for and are
a 1
b 3
a 1
a b 1 3
2
a 2b 1 6
5
The zeroes for and are
a 5
b 3
a 5
a b 5 3
2
a 2b 5 6
1
Thus, the values of and are and or and
a
b
1
3
5
and the zeroes of the given polynomial are ,
3
1
2
and .
5 Question: 3
Given that is a zero of the cubic polynomial
2
, find its other two zeroes.
3 2
6x 2x 10x 4 2
Solution:
Let
3 2
p x 6x 2x 10x 4 2
is one of the zeroes of .
2
p x
So, is one of the factors of given cubic
x 2
polynomial.
By long division,
Fig. 2.4.3
3 2
6x 2x 10x 4 2
2
6x 7 2x 4 x 2 0 2
6x 4 2x 3 2x 4 x 2
x 2 2x 3 2x 4 1 3 2x 4
x 2 3 2x 4 2x 1
x 2 3 2x 4 2x 1
Thus, the other zeroes are and
1
2
4
3 2
Questions: 4
Find so that is a factor of
k
2
x 2x k
. Also find all the zeroes of the
4 3 2
2x x 14x 5x 6
two polynomials.
Solution:
Apply long division,
Fig. 2.4.4 Since, is a factor of the given polynomial,
2
x 2x k
the remainder should be zero.
and
7k 21 0
2
2k 8k 6 0
and
k 3
2
k 4k 3 0
and
k 3
2
k 3k k 3 0
and
k 3
k k 3 1 k 3 0
and
k 3
k 3 k 1 0
and or
k 3
1 k
3 k
Only satisfies the required condition.
k 3
Now,
Dividend Divisor Quotient Remainder
4 3 2 2 2
2x x 14x 5x 6 x 2x 3 2x 3x 2 0
Factorisation by splitting method,
2 2
x 2x 3 2x 3x 2
2 2
x 3x x 3 2x 4x x 2
x x 3 1 x 3 2x x 2 1 x 2
x 3 x 1 x 2 2x 1
Thus, the zeroes of the polynomial are
2
2 3 x x
and are .
1, 3
4 3 2
2x x 14x 5x 6
1
1, 3,2,
2 Question: 5
Given that is a factor of the cubic polynomial
x 5
, find all the zeroes of the
3 2
x 3 5x 13x 3 5
polynomial.
Solution:
Apply long division,
Fig. 2.4.5
Now, find the zeroes by division algorithm.
3 2 2
x 3 5x 13x 3 5 x 2 5x 3 x 5 0
2
x 5 x 5 2 5 2 x 3
2
x 5 x 5 2 x 5 2 x 5 2 5 2
x 5 x x 5 2 5 2 x 5 2 x 5 x 5 2 x 5 2
Thus, all the zeroes of the polynomial are
and .
5, 5 2
5 2
Question: 6
For which values of a and b, are the zeroes of
also the zeroes of the polynomial
3 2
q x x 2x a
? Which zeroes of
5 4 3 2
p x x x 4x 3x 3x b
p(x) are not the zeroes of ?
q x
Solution:
Apply long division,
Fig. 2.4.6
Here, is a factor of the given
3 2
q x x 2x a
polynomial . So, the
5 4 3 2
p x x x 4x 3x 3x b
remainder should be zero.
2 2
1 a x 3 3a x b 2a 0.x 0.x 0
Compare the coefficients of and the constant
2
x
term.
a 1 0
a 1
b 2a 0
b 2a
b 2 1 2
Thus,
3 2
q x x 2x 1
Now,
3 2 2
p x x 2x 1 x 3x 2 0
3 2 2
x 2x 1 x 2x x 2
3 2
x 2x 1 x x 2 1 x 2
3 2
x 2x 1 x 2 x 1
Thus, and are the zeroes of but not the
1
2
p x
zeroes of .
q x