# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 2 – Polynomials

## Chapter 2 – Polynomials Exercise Ex. 2A

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

The given expression is an expression having only

non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 5. So, it is a polynomial of

degree 5.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

The given expression is an expression having only

non-negative integral powers of y. So, it is a polynomial.

The highest power of y is 3. So, it is a polynomial of

degree 3.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

The given expression is an expression having only

non-negative integral powers of t. So, it is a polynomial.

The highest power of t is 2. So, it is a polynomial of

degree 2.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

X^{100} – 1

X^{100} – 1

The given expression is an expression having only

non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 100. So, it is a polynomial

of degree 100.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

The given expression can be written as

It contains a term having negative integral power of

x. So, it is not a polynomial.

Which of the expressions are polynomials?

It is a polynomial, Degree = 2.

Which of the expressions are polynomials?

It is not a polynomial.

Which of the expressions are polynomials?

1

It is a polynomial, Degree = 0.

Which of the expressions are polynomials?

_{}

It is a polynomial, Degree = 0.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

The given expression is an expression having only

non-negative integral powers of x. So, it is a polynomial.

The highest power of x is 2. So, it is a polynomial of

degree 2.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

The given expression can be written as 2x^{-2}.

It contains a term having negative integral power of

x. So, it is not a polynomial.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

The given expression contains a term containing x^{1/2},

where ½ is not a non-negative integer.

So, it is not a polynomial.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

integral powers of x. So, it is a polynomial.

The highest power of x is 2. So, it is a polynomial of

degree 2.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

x^{4} – x^{3/2} + x – 3

x^{4} – x^{3/2} +

x – 3

The given expression contains a term containing x^{3/2},

where 3/2 is

not a non-negative integer.

So, it is not a polynomial.

Which of the following expressions are polynomials? In

case of a polynomial, write its degree.

The given expression can be written as 2x^{3}

+ 3x^{2} + x^{1/2} – 1.

The given expression contains a term containing x^{1/2},

where ½ is not a non-negative integer.

So, it is not a polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

–7 + x

-7

+ x

The

degree of a given polynomial is 1.

Hence,

it is a linear polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

6y

6y

The

degree of a given polynomial is 1.

Hence,

it is a linear polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

–z^{3}

-z^{3}

The

degree of a given polynomial is 3.

Hence,

it is a cubic polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

1 – y – y^{3}

1 – y – y^{3}

The

degree of a given polynomial is 3.

Hence,

it is a cubic polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

–p

–p

The

degree of a given polynomial is 1.

Hence,

it is a linear polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

x – x^{3} + x^{4}

x – x^{3} + x^{4}

The

degree of a given polynomial is 4.

Hence,

it is a quartic polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

1 + x + x^{2}

1 + x + x^{2}

The

degree of a given polynomial is 2.

Hence,

it is a quadratic polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

-6x^{2}

-6x^{2}

The

degree of a given polynomial is 2.

Hence,

it is a quadratic polynomial.

Identify constant, linear, quadratic, cubic and quartic

polynomials from the following.

–13

–13

The

given polynomial contains only one term namely constant.

Hence,

it is a constant polynomial.

Write the coefficient of x^{3} in x + 3x^{2}

– 5x^{3} + x^{4}

The coefficient of x^{3}

in x + 3x^{2} – 5x^{3} + x^{4} is -5.

Write the coefficient of x in .

The coefficient of x in .

Write the coefficient of x^{2} in 2x – 3 + x^{3}.

The given polynomial can be

written as x^{3} + 0x^{2} + 2x – 3.

Hence, the coefficient of x^{2}

in 2x – 3 + x^{3} is 0.

Write the coefficient of x in .

The coefficient of x in .

Write the constant term in .

The constant term in .

Determine the degree of each of the following polynomials.

Hence, the degree of a given polynomial is 2.

Determine the degree of each of the following polynomials.

y^{2}(y – y^{3})

y^{2}(y – y^{3})

= y^{3} – y^{5}

Hence, the degree of a given polynomial is 5.

Determine the degree of each of the following polynomials.

(3x –

2)(2x^{3} + 3x^{2})

(3x – 2)(2x^{3} + 3x^{2})

= 6x^{4} + 9x^{3}

– 4x^{3} – 6x^{2}

= 6x^{4} + 5x^{3}

– 6x^{2}

Hence, the degree of a given polynomial is 4.

Determine the degree of each of the following polynomials.

The degree of a given polynomial is 1.

Determine the degree of each of the following polynomials.

-8

-8

This

is a constant polynomial.

The

degree of a non-zero constant polynomial is zero.

Determine the degree of each of the following polynomials.

x^{-2}(x^{4} + x^{2})

x^{-2}(x^{4} + x^{2})

= x^{-2}.x^{2}(x^{2}

+ 1)

= x^{0} (x^{2} +

1)

= x^{2} + 1

Hence, the degree of a given

polynomial is 2.

Give an example of a monomial of degree 5.

Example of a monomial of degree 5:

3x^{5}^{ }

Give an example of a binomial of degree 8.

Example of a binomial of degree 8:

x – 6x^{8}^{ }

Give an example of a trinomial of degree 4.

Example of a trinomial of degree

4:

7 + 2y + y^{4}^{ }

Give an example of a monomial of degree 0.

Example of a monomial of degree 0:

7

Rewrite each of the following polynomials in standard

form.

x – 2x^{2} + 8 + 5x^{3}

x – 2x^{2} + 8 + 5x^{3}

in standard form:

5x^{3} – 2x^{2} +

x + 8

Rewrite each of the following polynomials in standard

form.

^{}

^{ }

^{ }

Rewrite each of the following polynomials in standard

form.

6x^{3} + 2x – x^{5} – 3x^{2}

6x^{3} + 2x – x^{5}

– 3x^{2} in standard form:

-x^{5} + 6x^{3} –

3x^{2} + 2x

Rewrite each of the following polynomials in standard

form.

2 + t – 3t^{3} + t^{4} – t^{2}

2 + t – 3t^{3} + t^{4}

– t^{2} in standard form:

t^{4} – 3t^{3} – t^{2}

+ t + 2

## Chapter 2 – Polynomials Exercise Ex. 2B

If p(x) = 5 – 4x + 2x^{2}, find

(i) p(0)

(ii) p(3)

(iii) p(-2)

p(x) = 5 – 4x + 2x^{2}

(i) p(0) = 5 – 4 _{} 0 + 2 _{} 0^{2} = 5

(ii) p(3) = 5 – 4 _{} 3 + 2 _{} 3^{2}

= 5 – 12 + 18

= 23 – 12 = 11

(iii) p(-2) = 5 – 4(-2) + 2 (-2)^{2}

= 5 + 8 + 8 = 21

If p(y) = 4 + 3y – y^{2} + 5y^{3}, find

(i) p(0)

(ii) p(2)

(iii) p(-1)

p(y) = 4 + 3y – y^{2} + 5y^{3}

(i) p(0) = 4 + 3 _{} 0 – 0^{2} + 5 _{} 0^{3}

= 4 + 0 – 0 + 0 = 4

(ii) p(2) = 4 + 3 _{} 2 – 2^{2} + 5 _{} 2^{3}

= 4 + 6 – 4 + 40

= 10 – 4 + 40 = 46

(iii) p(-1) = 4 + 3(-1) – (-1)^{2} + 5(-1)^{3}

= 4 – 3 – 1 – 5 = -5

If f(t) = 4t^{2} – 3t + 6, find

(i) f(0)

(ii) f(4)

(iii) f(-5)

f(t) = 4t^{2} – 3t + 6

(i) f(0) = 4 _{} 0^{2} – 3 _{} 0 + 6

= 0 – 0 + 6 = 6

(ii) f(4) = 4(4)^{2} – 3 _{} 4 + 6

= 64 – 12 + 6 = 58

(iii) f(-5) = 4(-5)^{2} – 3(-5) + 6

= 100 + 15 + 6 = 121

If p(x) = x^{3} – 3x^{2} + 2x, find p(0), p(1), p(2). What do you conclude?

p(x) = x^{3} – 3x^{2}

+ 2x

Thus, we have

p(0) = 0^{3} – 3(0)^{2}

+ 2(0) = 0

p(1) = 1^{3} – 3(1)^{2}

+ 2(1) = 1 – 3 + 2 = 0

p(2) = 2^{3} – 3(2)^{2}

+ 2(2) = 8 – 12 + 4 = 0

Hence, 0, 1 and 2 are the zeros of

the polynomial p(x) = x^{3} – 3x^{2} + 2x.

If p(x) = x^{3} + x^{2} – 9x – 9, find p(0), p(3), p(-3) and p(-1). What do you conclude about

the zeros of p(x)? Is 0 a zero of p(x)?

p(x) = x^{3} + x^{2}

– 9x – 9

Thus, we have

p(0) = 0^{3} + 0^{2}

– 9(0) – 9 = -9

p(3) = 3^{3} + 3^{2}

– 9(3) – 9 = 27 + 9 – 27 – 9 = 0

p(-3) = (-3)^{3} + (-3)^{2}

– 9(-3) – 9 = -27 + 9 + 27 – 9 = 0

p(-1) = (-1)^{3} + (-1)^{2}

– 9(-1) – 9 = -1 + 1 + 9 – 9 = 0

Hence, 0, 3 and -3 are the zeros

of p(x).

Now, 0 is not a zero of p(x) since

p(0) ≠ 0.

Verify that:

4 is a zero of the polynomial p(x) = x – 4.

p(x) = x – 4

Then, p(4) = 4 – 4 = 0

_{ }4 is a zero of the polynomial p(x).

Verify that:

-3 is a zero of the polynomial p(x) = x – 3.

p(x) = x – 3

Then,p(-3) = -3 – 3 = -6

-3 is not a zero of the polynomial p(x).

Verify that:

is a zero of the polynomial p(y) = 2y + 1.

p(y) = 2y + 1

Then, _{}

_{}is a zero of the polynomial p(y).

Verify that:

is a zero of thepolynomial p(x) = 2 – 5x.

p(x) = 2 – 5x

Then, _{}

_{ }is a zero of the polynomial p(x).

Verify that:

1 and 2 are the zeros of the polynomial p(x) = (x – 1) (x – 2).

p(x) = (x – 1) (x – 2)

Then,p(1)= (1 – 1) (1 – 2) = 0 _{}-1 = 0

_{ }1 is a zero of the polynomial p(x).

Also, p(2) = (2 – 1)(2 – 2) = 1 _{}0 = 0

_{ }2 is a zero of the polynomial p(x).

Hence,1 and 2 are the zeroes of the polynomial p(x).

Verify that:

2 and -3 are the zeros of the polynomial p(x) = x^{2} + x – 6.

p(x) = x^{2} + x – 6

Then, p(2) = 2^{2} + 2 – 6

= 4 + 2 – 6

= 6 – 6 = 0

_{ }2 is a zero of the polynomial p(x).

Also, p(-3) = (-3)^{2} – 3 – 6

= 9 – 3 – 6 = 0

_{ }-3 is a zero of the polynomial p(x).

Hence, 2 and -3 are the zeroes of the polynomial p(x).

Verify that:

0 and 3 are the zeros of the polynomial p(x) = x^{2} – 3x.

p(x) = x^{2} – 3x.

Then,p(0) = 0^{2} – 3 _{}0 = 0

p(3) = (3)^{2}– 3 _{}3 = 9 – 9 = 0

_{}0 and 3 are the zeroes of the polynomial p(x).

Find the zero of the polynomial:

p(x) = x – 5

p(x) = 0

x – 5 = 0

x = 5

_{5 is the zero of the polynomial p(x).}

Find the zero of the polynomial:

q(x) = x + 4

q(x) = 0

x + 4 = 0

x= -4

_{ -4 is the zero of the polynomial q(x).}

Find the zero of the polynomial:

r(x) = 2x + 5

r(x)

= 2x + 5

Now,

r(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

Find the zero of the polynomial:

f(x) = 3x + 1

f(x) = 0

3x + 1= 0

3x=-1

x =

x =is the zero of the polynomial f(x).

Find the zero of the polynomial:

g(x) = 5 – 4x

g(x) = 0

_{ }5 – 4x = 0

-4x = -5

x =_{}

x = is the zero of the polynomial g(x).

Find the zero of the polynomial:

h(x) = 6x – 2

h(x)

= 6x – 2

Now,

h(x) = 0

⇒ 6x – 2 = 0

⇒ 6x = 2

Find the zero of the polynomial:

p(x) = ax, a 0

p(x) = 0

_{ }ax = 0

x = 0

0 is the zero of the polynomial p(x).

Find the zero of the polynomial:

q(x) = 4x

q(x) = 0

4x = 0

x = 0

0 is the zero of the polynomial q(x).

If 2 and 0 are the zeros of the polynomial f(x) = 2x^{3}

– 5x^{2} + ax + b then find the values of a and

b.

HINT f(2) = 0 and f(0) = 0.

f(x) = 2x^{3} – 5x^{2}

+ ax + b

Now, 2 is a zero of f(x).

⇒

f(2) = 0

⇒

2(2)^{3} – 5(2)^{2} + a(2) + b = 0

⇒

16 – 20 + 2a + b = 0

⇒

2a + b – 4 = 0 ….(i)

Also, 0 is a zero of f(x).

⇒

f(0) = 0

⇒

2(0)^{3} – 5(0)^{2} + a(0) + b = 0

⇒

0 – 0 + 0 + b = 0

⇒

b = 0

Substituting b = 0 in (i), we get

2a + 0 – 4 = 0

⇒

2a = 4

⇒

a = 2

Thus, a = 2 and b = 0.

## Chapter 2 – Polynomials Exercise Ex. 2C

By actual division, find the quotient and the remainder

when (x^{4} + 1) is divided by (x – 1).

Verify that remainder = f(1).

Quotient = x^{3}

+ x^{2} + x + 1

Remainder = 2

Verification:

f(x) = x^{4} + 1

Then, f(1) = 1^{4}

+ 1 = 1 + 1 = 2 = Remainder

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = x^{3} – 6x^{2} + 2x – 4, g(x) = .

By

the remainder theorem, we know that when p(x) = x^{3} – 6x^{2} + 2x – 4 is divided by

g(x) = , the remainder is g.

Now,

Hence,

the required remainder is .

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = 2x^{3} + 3x^{2} – 11x – 3, g(x) = .

By

the remainder theorem, we know that when p(x) = 2x^{3} + 3x^{2} – 11x – 3 is divided by

g(x) = , the remainder is g.

Now,

Hence,

the required remainder is 3.

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = x^{3} – ax^{2} + 6x – a, g(x) = x –

a.

x

– a = 0

⇒ x = a

By

the remainder theorem, we know that when p(x) = x^{3} – ax^{2} + 6x – a is divided by

g(x) = x – a, the remainder is g(a).

Now,

g(a)

= (a)^{3}

– a(a)^{2} + 6(a)

– a

= a^{3}– a^{3}+ 6a – a = 5a

Hence,

the required remainder is 5a.

The polynomial (2x^{3} + x^{2} – ax + 2) and (2x^{3} – 3x^{2} – 3x + a)

when divided by (x – 2) leave the same remainder. Find the value of a.

Let p(x) = 2x^{3} + x^{2}

– ax + 2 and q(x) = 2x^{3} – 3x^{2}

– 3x + a be the given polynomials.

The remainders when p(x) and q(x)

are divided by (x – 2) are p(2) and q(2)

respectively.

By the given condition, we have

p(2) = q(2)

⇒

2(2)^{3} + (2)^{2} – a(2) + 2 = 2(2)^{3} – 3(2)^{2}

– 3(2) + a

⇒

16 + 4 – 2a + 2 = 16 – 12 – 6 + a

⇒

22 – 2a = -2 + a

⇒

a + 2a = 22 + 2

⇒

3a = 24

⇒

a = 8

The polynomial f(x) = x^{4} – 2x^{3 }+ 3x^{2} – ax + b when divided by (x – 1) and (x + 1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when f(x) is divided by (x – 2).

Letf(x) = (x^{4} – 2x^{3} + 3x^{2} – ax + b)

_{From the given information,}

f(1) = 1^{4} – 2(1)^{3} + 3(1)^{2} – a _{}1 + b = 5

1 – 2 + 3 – a + b = 5

2 – a + b = 5(i)

And,

f(-1) = (-1)^{4} – 2(-1)^{3} + 3(-1)^{2} – a(-1) + b = 19

1 + 2 + 3 + a + b = 19

6 + a + b = 19(ii)

Adding (i) and (ii), we get

8 + 2b = 24

_{}2b= 24 – 8 = 16

_{}b = _{}

Substituting the value of b = 8 in (i), we get

2 – a + 8 = 5

_{}-a + 10 = 5

_{}-a = -10 + 5

_{}-a = -5

_{}a = 5

_{}a = 5 and b = 8

f(x) = x^{4} – 2x^{3} + 3x^{2} – ax + b

= x^{4} – 2x^{3} + 3x^{2} – 5x + 8

_{}f(2) = (2)^{4} – 2(2)^{3} + 3(2)^{2} – 5 _{}2 + 8

= 16 – 16 + 12 – 10 + 8

= 20 – 10 = 10

_{}The required remainder is 10.

If p(x) = x^{3} – 5x^{2} + 4x – 3 and g(x)

= x – 2, show that p(x) is not a multiple of g(x).

The

polynomial p(x) will be a multiple of g(x) if g(x) divides p(x) completely.

i.e. when p(x) is

divided by g(x), it does not leave any remainder.

Now,

x – 2 = 0 ⇒ x = 2

Also,

p(2)

= (2)^{3} – 5(2)^{2} + 4(2) – 3 = 8 – 20 + 8 – 3 = -7 ≠ 0

Thus,

p(x) is not a multiple of g(x).

If p(x) = 2x^{3} – 11x^{2} – 4x + 5 and

g(x) = 2x + 1, show that g(x) is not a factor of p(x).

The

polynomial g(x) will be a factor of p(x) if g(x) divides p(x) completely.

i.e. when p(x) is

divided by g(x), it does not leave any remainder.

Now,

2x + 1 = 0 ⇒ x =

Also,

Thus,

g(x) is not a factor of p(x).

Verify the division algorithm for the polynomials

p(x) = 2x^{4} – 6x^{3} + 2x^{2} – x

+ 2 and g(x) = x + 2.

Using remainder theorem, find the remainder when:

(x^{3} – 6x^{2} + 9x + 3) is divided by (x – 1)

f(x) = x^{3} – 6x^{2} + 9x + 3

Now, x – 1 = 0 _{} x = 1

By the remainder theorem, we know that when f(x) is divided by (x – 1) the remainder is f(1).

Now, f(1) = 1^{3} – 6 _{} 1^{2} + 9 _{} 1 + 3

= 1 – 6 + 9 + 3

= 13 – 6 = 7

_{} The required remainder is 7.

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = 2x^{3} – 7x^{2} + 9x – 13, g(x) = x –

3.

x

– 3 = 0

⇒ x = 3

By

the remainder theorem, we know that when p(x) = 2x^{3} – 7x^{2} + 9x – 13 is divided by

g(x) = x – 3, the remainder is g(3).

Now,

g(3)

= 2(3)^{3} – 7(3)^{2}

+ 9(3) – 13

= 54 – 63 + 27 – 13 = 5

Hence,

the required remainder is 5.

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = 3x^{4} – 6x^{2} – 8x – 2, g(x) = x –

2.

x

– 2 = 0

⇒ x = 2

By

the remainder theorem, we know that when p(x) = 3x^{4} – 6x^{2} – 8x – 2 is divided by

g(x) = x – 2, the remainder is g(2).

Now,

g(2)

= 3(2)^{4} – 6(2)^{2}

– 8(2) – 2

= 48 – 24 – 16 – 2 = 6

Hence,

the required remainder is 6.

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = 2x^{3} – 9x^{2} + x + 15, g(x) = 2x –

3.

2x

– 3 = 0

⇒ x =

By

the remainder theorem, we know that when p(x) = 2x^{3} – 9x^{2} + x + 15 is divided by

g(x) = 2x – 3, the remainder is g.

Now,

Hence,

the required remainder is 3.

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = x^{3} – 2x^{2} – 8x – 1, g(x) = x + 1.

x

+ 1 = 0

⇒ x = -1

By

the remainder theorem, we know that when p(x) = x^{3} – 2x^{2} – 8x – 1 is divided by

g(x) = x + 1, the remainder is g(-1).

Now,

g(-1)

= (-1)^{3} – 2(-1)^{2} – 8(-1) – 1 = -1 – 2 + 8 – 1 = 4

Hence,

the required remainder is 4.

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = 2x^{3} + x^{2}– 15x – 12, g(x) = x + 2.

x

+ 2 = 0

⇒ x = -2

By

the remainder theorem, we know that when p(x) = 2x^{3} + x^{2} – 15x – 12 is divided by

g(x) = x + 2, the remainder is g(-2).

Now,

g(-2)

= 2(-2)^{3} + (-2)^{2} – 15(-2) – 12 = -16 + 4 + 30 – 12 = 6

Hence,

the required remainder is 6.

Using the remainder theorem, find the remainder, when p(x)

is divided by g(x), where

p(x) = 6x^{3} + 13x^{2} + 3, g(x) = 3x + 2.

3x

+ 2 = 0

⇒ x =

By

the remainder theorem, we know that when p(x) = 6x^{3} + 13x^{2} + 3 is divided by

g(x) = 3x + 2, the remainder is g.

Now,

Hence,

the required remainder is 7.

## Chapter 2 – Polynomials Exercise Ex. 2D

Using factor theorem, show that:

(x – 2) is a factor of (x^{3} – 8)

f(x) = (x^{3} – 8)

By the Factor Theorem, (x – 2) will be a factor of f(x) if f(2) = 0.

Here, f(2) = (2)^{3} – 8

= 8 – 8 = 0

_{} (x – 2) is a factor of (x^{3} – 8).

Using factor theorem, show that:

(x + _{}) is a factor of _{}

f(x) = _{}

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here,

_{}

Show that (p – 1) is a factor of (p^{10} – 1) and

also of (p^{11} – 1).

Let q(p) = (p^{10} – 1)

and f(p) = (p^{11} – 1)

By the factor theorem, (p – 1)

will be a factor of q(p) and f(p) if q(1) and f(1) =

0.

Now, q(p) = p^{10} – 1

⇒

q(1) = 1^{10} – 1 = 1 – 1 = 0

Hence, (p – 1) is a factor of p^{10}

– 1.

And, f(p) = p^{11} – 1

⇒

f(1) = 1^{11} – 1 = 1 – 1 = 0

Hence, (p – 1) is also a factor of

p^{11} – 1.

Find the value of k for which (x – 1) is a factor of (2x^{3}+ 9x^{2} + x + k).

f(x) = (2x^{3} + 9x^{2} + x + k)

x – 1 = 0 _{} x = 1

_{} f(1) = 2 _{} 1^{3} + 9 _{} 1^{2} + 1 + k

= 2 + 9 + 1 + k

= 12 + k

Given that (x – 1) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(1) = 0.

f(1) = 12 + k = 0

k = -12.

Find the value of a for which (x – 4) is a factor of (2x^{3} – 3x^{2} – 18x + a).

f(x) = (2x^{3} – 3x^{2} – 18x + a)

x – 4 = 0 _{} x = 4

_{} f(4) = 2(4)^{3} – 3(4)^{2} – 18 _{} 4 + a

= 128 – 48 – 72 + a

= 128 – 120 + a

= 8 + a

Given that (x – 4) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(4) = 0.

f(4) = 8 + a = 0

a = -8

Find the value of a for which (x

+ 1) is a factor of (ax^{3} + x^{2} – 2x + 4a – 9).

Let

p(x) = ax^{3} + x^{2} –

2x + 4a – 9

It is given that (x + 1) is a

factor of p(x).

⇒

p(-1) = 0

⇒

a(-1)^{3} + (-1)^{2} – 2(-1) + 4a – 9 = 0

⇒ -a

+ 1 + 2 + 4a – 9 = 0

⇒

3a – 6 = 0

⇒

3a = 6

⇒

a = 2

Find the value of a for which (x

+ 2a) is a factor of (x^{5} – 4a^{2}x^{3} + 2x + 2a +

3).

Let

p(x) = x^{5} – 4a^{2}x^{3}

+ 2x + 2a + 3

It is given that (x + 2a) is a

factor of p(x).

⇒

p(-2a) = 0

⇒

(-2a)^{5} – 4a^{2}(-2a)^{3} + 2(-2a) + 2a + 3 = 0

⇒ -32a^{5}

– 4a^{2}(-8a^{3}) – 4a + 2a + 3 = 0

⇒

-32a^{5} + 32a^{5} -2a + 3 = 0

⇒

2a = 3

Find the value of m for which (2x – 1) is a factor of (8x^{4}

+ 4x^{3} – 16x^{2} + 10x + m).

Let

p(x) = 8x^{4} + 4x^{3} –

16x^{2} + 10x + m

It is given that (2x – 1) is a

factor of p(x).

Find the value of a for which the

polynomial (x^{4} – x^{3} – 11x^{2} – x + a) is

divisible by (x + 3).

Let

p(x) = x^{4} – x^{3} –

11x^{2} – x + a

It is given that p(x) is divisible

by (x + 3).

⇒

(x + 3) is a factor of p(x).

⇒

p(-3) = 0

⇒

(-3)^{4} – (-3)^{3} – 11(-3)^{2} – (-3) + a = 0

⇒ 81 + 27 – 99 + 3 + a = 0

⇒

12 + a = 0

⇒

a = -12

Without

actual division, show that (x^{3} – 3x^{2} – 13x + 15) is

exactly divisible by (x^{2} + 2x – 3).

Let f(x) = x^{3}

– 3x^{2} – 13x + 15

Now, x^{2} + 2x – 3 = x^{2}

+ 3x – x – 3

= x (x + 3) – 1

(x + 3)

= (x +

3) (x – 1)

Thus,

f(x) will be exactly divisible by x^{2} + 2x – 3 = (x + 3) (x – 1) if

(x + 3) and (x – 1) are both factors of f(x), so by factor theorem, we should

have f(-3) = 0 and f(1) = 0.

Now, f(-3) = (-3)^{3} – 3 (-3)^{2}

– 13 (-3) + 15

= -27 – 3 _{} 9 + 39 + 15

= -27 – 27 +

39 + 15

= -54 + 54 = 0

And, f(1) = 1^{3} – 3 _{} 1^{2} –

13 _{} 1 + 15

= 1 – 3 –

13 + 15

= 16 – 16 =

0

_{} f(-3) = 0 and f(1) = 0

So,

x^{2} + 2x – 3 divides f(x) exactly.

If (x^{3} + ax^{2} + bx + 6) has (x – 2) as a factor and leaves a remainder 3 when divided by (x – 3), find the values of a and b.

Letf(x) = (x^{3} + ax^{2} + bx + 6)

Now, by remainder theorem, f(x) when divided by (x – 3) will leave a remainder as f(3).

So, f(3) = 3^{3} + a _{}3^{2} + b _{}3 + 6 = 3

_{}27 + 9a + 3b + 6 = 3

_{}9 a + 3b + 33 = 3

_{}9a + 3b = 3 – 33

_{}9a + 3b = -30

_{}3a + b = -10(i)

Given that (x – 2) is a factor of f(x).

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(2) = 0.

_{} f(2) = 2^{3} + a _{}2^{2} + b _{}2 + 6 = 0

_{} 8 + 4a+ 2b + 6 = 0

_{ }4a + 2b = -14

_{} 2a + b = -7(ii)

Subtracting (ii) from (i), we get,

_{}a = -3

Substituting the value of a = -3 in (i), we get,

3(-3) + b = -10

_{}-9 + b = -10

_{}b = -10 + 9

_{}b = -1

_{}a = -3 and b = -1.

Using factor theorem, show that:

(x – 3) is a factor of (2x^{3} + 7x^{2} – 24x – 45)

f(x) = (2x^{3} + 7x^{2} – 24x – 45)

By the Factor Theorem, (x – 3) will be a factor of f(x) if f(3) = 0.

Here, f(3) = 2 _{} 3^{3} + 7 _{} 3^{2} – 24 _{} 3 – 45

= 54 + 63 – 72 – 45

= 117 – 117 = 0

_{} (x – 3) is a factor of (2x^{3} + 7x^{2} – 24x – 45).

Find the values of a and b so that the polynomial (x^{3} – 10x^{2} + ax + b) is exactly divisible by (x – 1) as well as (x – 2).

Let f(x) = (x^{3} – 10x^{2} + ax + b), then by factor theorem

(x – 1) and (x – 2) will be factors of f(x) if f(1) = 0 and f(2) = 0.

_{}f(1) = 1^{3} – 10 _{}1^{2} + a _{}1 + b = 0

_{}1 – 10 + a + b = 0

_{}a + b = 9(i)

Andf(2) = 2^{3} – 10 _{}2^{2} + a _{}2 + b = 0

_{}8 – 40 + 2a + b = 0

_{}2a + b = 32(ii)

Subtracting (i) from (ii), we get

a = 23

Substituting the value of a = 23 in (i), we get

23 + b = 9

_{}b = 9 – 23

_{}b = -14

_{}a = 23 and b = -14.

Find the values of a and b so that the polynomial (x^{4} + ax^{3} – 7x^{2} – 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

Letf(x)= (x^{4} + ax^{3} – 7x^{2} – 8x + b)

Now, x + 2 = 0 _{}x = -2 and x + 3 = 0 _{}x = -3

By factor theorem, (x + 2) and (x + 3) will be factors of f(x) if f(-2) = 0 and f(-3) = 0

_{}f(-2) = (-2)^{4} + a (-2)^{3} – 7 (-2)^{2} – 8 (-2) + b = 0

_{}16 – 8a – 28 + 16 + b = 0

_{}-8a + b = -4

_{}8a – b = 4(i)

And, f(-3) = (-3)^{4} + a (-3)^{3} – 7 (-3)^{2} – 8 (-3) + b = 0

_{}81 – 27a – 63 + 24 + b = 0

_{}-27a + b = -42

_{}27a – b = 42(ii)

Subtracting (i) from (ii), we get,

19a = 38

So, a = 2

Substituting the value of a = 2 in (i), we get

8 _{}2 – b = 4

_{}16 – b = 4

_{}-b = -16 + 4

_{}-b = -12

_{}b = 12

_{}a = 2 and b = 12.

If both (x – 2) and are factors of px^{2}

+ 5x + r, prove that p = r.

Let

f(x) = px^{2} + 5x + r

Now,

(x – 2) is a factor of f(x).

⇒ f(2) = 0

⇒ p(2)^{2} + 5(2) + r = 0

⇒ 4p + 10 + r = 0

⇒ 4p + r = -10

Also,

is a factor of

f(x).

From

(i) and (ii), we have

4p

+ r = p + 4r

⇒ 4p – p = 4r – r

⇒ 3p = 3r

⇒ p = r

Without actual division, prove that 2x^{4} – 5x^{3}

+ 2x^{2} – x + 2 is divisible by x^{2} – 3x + 2.

Let f(x) = 2x^{4} – 5x^{3}

+ 2x^{2} – x + 2

and g(x) = x^{2} – 3x + 2

=

x^{2} – 2x – x + 2

=

x(x – 2) – 1(x – 2)

=

(x – 2)(x – 1)

Clearly, (x – 2) and (x – 1) are

factors of g(x).

In order to prove that f(x) is

exactly divisible by g(x), it is sufficient to prove that f(x) is exactly

divisible by (x – 2) and (x – 1).

Thus, we will show that (x – 2)

and (x – 1) are factors of f(x).

Now,

f(2) = 2(2)^{4} – 5(2)^{3}

+ 2(2)^{2} – 2 + 2 = 32 – 40 + 8 = 0 and

f(1) = 2(1)^{4} – 5(1)^{3}

+ 2(1)^{2} – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0

Therefore, (x – 2) and (x – 1) are

factors of f(x).

⇒

g(x) = (x – 2)(x – 1) is a factor of f(x).

Hence, f(x) is exactly divisible

by g(x).

What must be added to 2x^{4} – 5x^{3} + 2x^{2}

– x – 3 so that the result is exactly divisible by (x – 2)?

Let the required number to be

added be k.

Then, p(x) = 2x^{4} – 5x^{3}

+ 2x^{2} – x – 3 + k and g(x) = x – 2

Thus, we have

p(2) = 0

⇒

2(2)^{4} – 5(2)^{3} + 2(2)^{2} – 2 – 3 + k = 0

⇒

32 – 40 + 8 – 5 + k = 0

⇒

k – 5 = 0

⇒

k = 5

Hence, the required number to be

added is 5.

What must be subtracted from (x^{4} + 2x^{3}

– 2x^{2} + 4x + 6) so that the result is exactly divisible by (x^{2}

+ 2x – 3)?

Let p(x) = x^{4} + 2x^{3}

– 2x^{2} + 4x + 6 and q(x) = x^{2} + 2x – 3.

When p(x) is divided by q(x), the

remainder is a linear expression in x.

So, let r(x) = ax + b be subtracted

from p(x) so that p(x) – r(x) is divided by q(x).

Let f(x) = p(x) – r(x) = p(x) –

(ax + b)

= (x^{4} + 2x^{3}

– 2x^{2} + 4x + 6) – (ax + b)

= x^{4} + 2x^{3} –

2x^{2} + (4 – a)x + 6 – b

We have,

q(x) = x^{2} + 2x – 3

=

x^{2} + 3x – x – 3

=

x(x + 3) – 1(x + 3)

=

(x + 3)(x – 1)

Clearly, (x + 3) and (x – 1) are

factors of q(x).

Therefore, f(x) will be divisible

by q(x) if (x + 3) and (x – 1) are factors of f(x).

i.e., f(-3) = 0 and f(1) = 0

Consider, f(-3) = 0

⇒

(-3)^{4} + 2(-3)^{3} – 2(-3)^{2} + (4 – a)(-3) + 6 –

b = 0

⇒

81 – 54 – 18 – 12 + 3a + 6 – b = 0

⇒

3 + 3a – b = 0

⇒

3a – b = -3 ….(i)

And, f(1) = 0

⇒

(1)^{4} + 2(1)^{3} – 2(1)^{2} + (4 – a)(1) + 6 – b =

0

⇒

1 + 2 – 2 + 4 – a + 6 – b = 0

⇒

11 – a – b = 0

⇒

-a – b = -11 ….(ii)

Subtracting (ii) from (i), we get

4a = 8

⇒

a = 2

Substituting a = 2 in (i), we get

3(2) – b = -3

⇒

6 – b = -3

⇒

b = 9

Putting the values of a and b in

r(x) = ax + b, we get

r(x) = 2x + 9

Hence, p(x) is divisible by q(x),

if r(x) = 2x + 9 is subtracted from it.

Use factor theorem to prove that (x + a) is a factor of (x^{n} + a^{n}) for any odd positive

integer n.

Let f(x) = x^{n}

+ a^{n}

In order to prove that (x + a) is

a factor of f(x) for any odd positive integer n, it is sufficient to show

that f(-a) = 0.

Now,

f(-a) = (-a)^{n} + a^{n}

=

(-1)^{n} a^{n} + a^{n}

=

[(-1)^{n} + 1] a^{n}

=

[-1 + 1] a^{n} …[n is odd ⇒

(-1)^{n} = -1]

=

0 ×

a^{n}

=

0

Hence, (x + a) is a factor of x^{n} + a^{n} for any odd positive

integer n.

Using factor theorem, show that:

(x – 1) is a factor of (2x^{4} + 9x^{3} + 6x^{2} – 11x – 6)

f(x) = (2x^{4} + 9x^{3} + 6x^{2} – 11x – 6)

By the Factor Theorem, (x – 1) will be a factor of f(x) if f(1) = 0.

Here, f(1) = 2 _{} 1^{4} + 9 _{} 1^{3} + 6 _{} 1^{2} – 11 _{} 1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17 = 0

_{} (x – 1) is factor of (2x^{4} + 9x^{3} + 6x^{2} – 11x – 6).

Using factor theorem, show that:

(x + 2) is a factor of (x^{4} – x^{2} – 12)

f(x) = (x^{4} – x^{2} – 12)

By the Factor Theorem, (x + 2) will be a factor of f(x) if f(-2) = 0.

Here, f(-2) = (-2)^{4} – (-2)^{2} – 12

= 16 – 4 – 12

= 16 – 16 = 0

_{} (x + 2) is a factor of (x^{4} – x^{2} – 12).

p(x) = 69 + 11x – x^{2} + x^{3}, g(x) = x

+ 3

By

the factor theorem, g(x) = x + 3 will be a factor of p(x) if p(-3) = 0.

Now, p(x) = 69 + 11x – x^{2}

+ x^{3}

⇒ p(-3) = 69 + 11(-3) – (-3)^{2} + (-3)^{3}

=

69 – 33 – 9 – 27

=

0

Hence, g(x) = x + 3 is a factor of

the given polynomial p(x).

Using factor theorem, show that:

(x + 5) is a factor of (2x^{3} + 9x^{2} – 11x – 30)

f(x) = 2x^{3} + 9x^{2} – 11x – 30

By the Factor Theorem, (x + 5) will be a factor of f(x) if f(-5) = 0.

Here, f(-5) = 2(-5)^{3} + 9(-5)^{2} – 11(-5) – 30

= -250 + 225 + 55 – 30

= -280 + 280 = 0

_{} (x + 5) is a factor of (2x^{3} + 9x^{2} – 11x – 30).

Using factor theorem, show that:

(2x – 3) is a factor of (2x^{4} + x^{3} – 8x^{2} – x + 6)

f(x) = (2x^{4} + x^{3} – 8x^{2} – x + 6)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here, 2x – 3 = 0 _{} x = _{}

_{} _{}

_{}

_{}

_{} is a factor of _{}.

p(x) = 3x^{3} + x^{2} – 20x + 12, g(x) =

3x – 2

By

the factor theorem, g(x) = 3x – 2 will be a factor of p(x) if = 0.

Now, p(x) = 3x^{3} + x^{2}

– 20x + 12

Hence, g(x) = 3x – 2 is a factor

of the given polynomial p(x).

Using factor theorem, show that:

(x – _{}) is a factor of _{}

f(x) = _{}

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here,

= 14 – 8 – 6

= 14 – 14 = 0

_{}

## Chapter 2 – Polynomials Exercise MCQ

Which of the following

expressions is a polynomial in one variable?

Degree of the zero

polynomial is

(a) 1

(b) 0

(c) not defined

(d)none of these

Zero of the zero

polynomial is

(a) 0

(b) 1

(c) every real

number

(d)not defined

If p(x) = x + 4, then

p(x) + p(-x) = ?

(a) 0

(b) 4

(c) 2x

(d)8

If p(x) = 5x – 4x^{2} + 3 then p(-1) = ?

(a) 2

(b) -2

(c) 6

(d) -6

Correct

option: (d)

P(x)

= 5x – 4x^{2} + 3

⇒ p(-1) = 5(-1) – 4(-1)^{2}

+ 3 = -5 – 4 + 3 = -6

If (x^{51} + 51) is divided by (x + 1) then the

remainder is

(a) 0

(b) 1

(c) 49

(d) 50

Correct

option: (d)

Let

f(x) = x^{51} + 51

By

the remainder theorem, when f(x) is divided by (x + 1), the remainder is f(-1).

Now,

f(-1) = [(-1)^{n} + 51] = -1 + 51 = 50

If (x + 1) is a factor of the polynomial (2x^{2} + kx) then k = ?

(a) 4

(b) -3

(c) 2

(d) -2

Correct option: (c)

Let p(x) = 2x^{2} + kx

Since (x + 1) is a factor of p(x),

P(–1) = 0

⇒ 2(–1)^{2} + k(–1) = 0

⇒ 2 – k = 0

⇒ k = 2

When p(x) = x^{4}

+ 2x^{3} – 3x^{2} + x – 1 is divided by (x – 2), the

remainder is

(a) 0

(b) -1

(c) -15

(d)21

When p(x) = x^{3}

– 3x^{2 }+ 4x + 32 is divided

by (x + 2), the remainder is

(a) 0

(b) 32

(c) 36

(d)4

When p(x) = 4x^{3}

– 12x^{2} + 11x – 5 is divided by (2x – 1), the remainder is

(a) 0

(b) -5

(c) -2

(d)2

Which of the following

expressions is a polynomial?

When p(x) =x^{3}-ax^{2}+x

is divided by (x-a), the remainder is

(a) 0

(b) a

(c) 2a

(d)3a

When p(x) = x^{3}

+ ax^{2} + 2x + a is divided by (x + a), the remainder is

(a) 0

(b) a

(c) -a

(d)2a

(x + 1) is a factor of

the polynomial

(a) x^{3} –

2x^{2} + x + 2

(b) x^{3} +

2x^{2} + x – 2

(c) x^{3} +

2x^{2} – x – 2

(d)x^{3} +

2x^{2} – x + 2

Zero of the polynomial p(x) = 2x + 5 is

(a)

(b)

(c)

(d)

Correct

option: (b)

p(x)

= 2x + 5

Now,

p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

The zeroes of the

polynomial p(x) = x^{2} + x – 6 are

(a) 2, 3

(b) -2, 3

(c) 2, -3

(d)-2, -3

The zeroes of the

polynomial p(x) = 2x^{2} + 5x – 3 are

The zeros of the polynomial p(x) = 2x^{2} + 7x – 4

are

(a)

(b)

(c)

(d)

Correct

option: (c)

p(x)

= 2x^{2} + 7x – 4

Now,

p(x) = 0

⇒ 2x^{2} + 7x – 4 = 0

⇒ 2x^{2} + 8x – x – 4 = 0

⇒ 2x(x + 4) – 1(x + 4) = 0

⇒ (x + 4)(2x – 1) = 0

⇒ x + 4 = 0 and 2x – 1 = 0

⇒ x = -4 and x =

If (x + 5) is a factor

of p(x) = x^{3} – 20x + 5k, then k =?

(a) -5

(b) 5

(c) 3

(d)-3

If (x + 2) and (x – 1)

are factors of (x^{3} + 10x^{2} + mx

+ n), then

(a) m = 5, n = -3

(b) m = 7, n = -18

(c) m = 17, n = -8

(d)m = 23, n = -19

If (x^{100} + 2x^{99}

+ k) is divisible by (x + 1), then the value of k is

(a) 1

(b) 2

(c) -2

(d)-3

Which of the following

is a polynomial?

For what value of k is

the polynomial p(x) = 2x^{3} – kx^{2} + 3x + 10 exactly

divisible by (x + 2)?

The zeroes of the polynomial

p(x) = x^{2} – 3x are

(a) 0, 0

(b) 0, 3

(c) 0, -3

(d)3, -3

The zeros of the polynomial p(x) = 3x^{2} – 1 are

(a)

(b)

(c)

(d)

Correct

option: (d)

p(x)

= 3x^{2} – 1

Now,

p(x) = 0

⇒ 3x^{2 }–

1 = 0

⇒ 3x^{2}

= 1

Which of the following

is a polynomial?

Which of the following

is a polynomial?

(a) x^{-2}

+ x^{-1} + 3

(b) x + x^{-1}

+ 2

(c) x^{-1}

(d)0

Which of the following

is a quadratic polynomial?

(a) x + 4

(b) x^{3} +

x

(c) x^{3} +

2x + 6

(d)x^{2} +

5x + 4

Which of the following

is a linear polynomial?

(a) x + x^{2}

(b) x + 1

(c) 5x^{2}

– x + 3

(d)

Which of the following

is a binomial?

(a) x^{2} + x + 3

(b) x^{2 }+ 4

(c) 2x^{2}

(d)

(a)

(b) 2

(c) 1

(d)0