Lesson: Polynomials

EXERCISE 2.1 (Multiple Choice Questions and

Answers)

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

p 3 0

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

p 3 0

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

quadratic polynomial?

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

Answer the following and justify:

(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

your answers:

(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