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

## Chapter 2 – Polynomials Exercise Ex. 2A

Question 1

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 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 5. So, it is a polynomial of
degree 5.

Question 2

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 2

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.

Question 3

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 3

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.

Question 4

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

X100 – 1

Solution 4

X100 – 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.

Question 5

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 5

The given expression can be written as

It contains a term having negative integral power of
x. So, it is not a polynomial.

Question 6

Which of the expressions are polynomials?

Solution 6

It is a polynomial, Degree = 2.

Question 7

Which of the expressions are polynomials?

Solution 7

It is not a polynomial.

Question 8

Which of the expressions are polynomials?

1

Solution 8

It is a polynomial, Degree = 0.

Question 9

Which of the expressions are polynomials?

Solution 9

It is a polynomial, Degree = 0.

Question 10

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 10

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.

Question 11

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 11

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.

Question 12

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 12

The given expression contains a term containing x1/2,
where ½ is not a non-negative integer.

So, it is not a polynomial.

Question 13

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 13

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.

Question 14

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

x4 – x3/2 + x – 3

Solution 14

x4 – x3/2 +
x – 3

The given expression contains a term containing x3/2,
where 3/2 is
not a non-negative integer.

So, it is not a polynomial.

Question 15

Which of the following expressions are polynomials? In
case of a polynomial, write its degree.

Solution 15

The given expression can be written as 2x3
+ 3x2 + x1/2 – 1.

The given expression contains a term containing x1/2,
where ½ is not a non-negative integer.

So, it is not a polynomial.

Question 16

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

7 + x

Solution 16

-7
+ x

The
degree of a given polynomial is 1.

Hence,
it is a linear polynomial.

Question 17

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

6y

Solution 17

6y

The
degree of a given polynomial is 1.

Hence,
it is a linear polynomial.

Question 18

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

z3

Solution 18

-z3

The
degree of a given polynomial is 3.

Hence,
it is a cubic polynomial.

Question 19

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

1 – y – y3

Solution 19

1 – y – y3

The
degree of a given polynomial is 3.

Hence,
it is a cubic polynomial.

Question 20

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

p

Solution 20

p

The
degree of a given polynomial is 1.

Hence,
it is a linear polynomial.

Question 21

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

x – x3 + x4

Solution 21

x – x3 + x4

The
degree of a given polynomial is 4.

Hence,
it is a quartic polynomial.

Question 22

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

1 + x + x2

Solution 22

1 + x + x2

The
degree of a given polynomial is 2.

Hence,

Question 23

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

-6x2

Solution 23

-6x2

The
degree of a given polynomial is 2.

Hence,

Question 24

Identify constant, linear, quadratic, cubic and quartic
polynomials from the following.

13

Solution 24

13

The
given polynomial contains only one term namely constant.

Hence,
it is a constant polynomial.

Question 25

Write the coefficient of x3 in x + 3x2
– 5x3 + x4

Solution 25

The coefficient of x3
in x + 3x2 – 5x3 + x4 is -5.

Question 26

Write the coefficient of x in .

Solution 26

The coefficient of x in .

Question 27

Write the coefficient of x2 in 2x – 3 + x3.

Solution 27

The given polynomial can be
written as x3 + 0x2 + 2x – 3.

Hence, the coefficient of x2
in 2x – 3 + x3 is 0.

Question 28

Write the coefficient of x in .

Solution 28

The coefficient of x in .

Question 29

Write the constant term in .

Solution 29

The constant term in .

Question 30

Determine the degree of each of the following polynomials.

Solution 30

Hence, the degree of a given polynomial is 2.

Question 31

Determine the degree of each of the following polynomials.

y2(y – y3)

Solution 31

y2(y – y3)

= y3 – y5

Hence, the degree of a given polynomial is 5.

Question 32

Determine the degree of each of the following polynomials.

(3x
2)(2x3 + 3x2)

Solution 32

(3x – 2)(2x3 + 3x2)

= 6x4 + 9x3
– 4x3 – 6x2

= 6x4 + 5x3
– 6x2

Hence, the degree of a given polynomial is 4.

Question 33

Determine the degree of each of the following polynomials.

Solution 33

The degree of a given polynomial is 1.

Question 34

Determine the degree of each of the following polynomials.

-8

Solution 34

-8

This
is a constant polynomial.

The
degree of a non-zero constant polynomial is zero.

Question 35

Determine the degree of each of the following polynomials.

x-2(x4 + x2)

Solution 35

x-2(x4 + x2)

= x-2.x2(x2
+ 1)

= x0 (x2 +
1)

= x2 + 1

Hence, the degree of a given
polynomial is 2.

Question 36

Give an example of a monomial of degree 5.

Solution 36

Example of a monomial of degree 5:

3x5

Question 37

Give an example of a binomial of degree 8.

Solution 37

Example of a binomial of degree 8:

x – 6x8

Question 38

Give an example of a trinomial of degree 4.

Solution 38

Example of a trinomial of degree
4:

7 + 2y + y4

Question 39

Give an example of a monomial of degree 0.

Solution 39

Example of a monomial of degree 0:

7

Question 40

Rewrite each of the following polynomials in standard
form.

x – 2x2 + 8 + 5x3

Solution 40

x – 2x2 + 8 + 5x3
in standard form:

5x3 – 2x2 +
x + 8

Question 41

Rewrite each of the following polynomials in standard
form.

Solution 41

Question 42

Rewrite each of the following polynomials in standard
form.

6x3 + 2x – x5 – 3x2

Solution 42

6x3 + 2x – x5
– 3x2 in standard form:

-x5 + 6x3
3x2 + 2x

Question 43

Rewrite each of the following polynomials in standard
form.

2 + t – 3t3 + t4 – t2

Solution 43

2 + t – 3t3 + t4
– t2 in standard form:

t4 – 3t3 – t2
+ t + 2

## Chapter 2 – Polynomials Exercise Ex. 2B

Question 1

If p(x) = 5 – 4x + 2x2, find

(i) p(0)

(ii) p(3)

(iii) p(-2)

Solution 1

p(x) = 5 – 4x + 2x2

(i) p(0) = 5 – 4 0 + 2 02 = 5

(ii) p(3) = 5 – 4 3 + 2 32

= 5 – 12 + 18

= 23 – 12 = 11

(iii) p(-2) = 5 – 4(-2) + 2 (-2)2

= 5 + 8 + 8 = 21

Question 2

If p(y) = 4 + 3y – y2 + 5y3, find

(i) p(0)

(ii) p(2)

(iii) p(-1)

Solution 2

p(y) = 4 + 3y – y2 + 5y3

(i) p(0) = 4 + 3 0 – 02 + 5 03

= 4 + 0 – 0 + 0 = 4

(ii) p(2) = 4 + 3 2 – 22 + 5 23

= 4 + 6 – 4 + 40

= 10 – 4 + 40 = 46

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

= 4 – 3 – 1 – 5 = -5

Question 3

If f(t) = 4t2 – 3t + 6, find

(i) f(0)

(ii) f(4)

(iii) f(-5)

Solution 3

f(t) = 4t2 – 3t + 6

(i) f(0) = 4 02 – 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

Question 4

If p(x) = x3 – 3x2 + 2x, find p(0), p(1), p(2). What do you conclude?

Solution 4

p(x) = x3 – 3x2
+ 2x

Thus, we have

p(0) = 03 – 3(0)2
+ 2(0) = 0

p(1) = 13 – 3(1)2
+ 2(1) = 1 – 3 + 2 = 0

p(2) = 23 – 3(2)2
+ 2(2) = 8 – 12 + 4 = 0

Hence, 0, 1 and 2 are the zeros of
the polynomial p(x) = x3 – 3x2 + 2x.

Question 5

If p(x) = x3 + x2 – 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)?

Solution 5

p(x) = x3 + x2
– 9x – 9

Thus, we have

p(0) = 03 + 02
– 9(0) – 9 = -9

p(3) = 33 + 32
– 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.

Question 6

Verify that:

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

Solution 6

p(x) = x – 4

Then, p(4) = 4 – 4 = 0

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

Question 7

Verify that:

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

Solution 7

p(x) = x – 3

Then,p(-3) = -3 – 3 = -6

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

Question 8

Verify that:

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

Solution 8

p(y) = 2y + 1

Then,

is a zero of the polynomial p(y).

Question 9

Verify that:

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

Solution 9

p(x) = 2 – 5x

Then,

is a zero of the polynomial p(x).

Question 10

Verify that:

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

Solution 10

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).

Question 11

Verify that:

2 and -3 are the zeros of the polynomial p(x) = x2 + x – 6.

Solution 11

p(x) = x2 + x – 6

Then, p(2) = 22 + 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).

Question 12

Verify that:

0 and 3 are the zeros of the polynomial p(x) = x2 – 3x.

Solution 12

p(x) = x2 – 3x.

Then,p(0) = 02 – 3 0 = 0

p(3) = (3)2– 3 3 = 9 – 9 = 0

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

Question 13

Find the zero of the polynomial:

p(x) = x – 5

Solution 13

p(x) = 0

x – 5 = 0

x = 5

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

Question 14

Find the zero of the polynomial:

q(x) = x + 4

Solution 14

q(x) = 0

x + 4 = 0

x= -4

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

Question 15

Find the zero of the polynomial:

r(x) = 2x + 5

Solution 15

r(x)
= 2x + 5

Now,
r(x) = 0

2x + 5 = 0

2x = -5

Question 16

Find the zero of the polynomial:

f(x) = 3x + 1

Solution 16

f(x) = 0

3x + 1= 0

3x=-1

x =

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

Question 17

Find the zero of the polynomial:

g(x) = 5 – 4x

Solution 17

g(x) = 0

5 – 4x = 0

-4x = -5

x =

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

Question 18

Find the zero of the polynomial:

h(x) = 6x – 2

Solution 18

h(x)
= 6x
– 2

Now,
h(x) = 0

6x – 2 = 0

6x = 2

Question 19

Find the zero of the polynomial:

p(x) = ax, a  0

Solution 19

p(x) = 0

ax = 0

x = 0

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

Question 20

Find the zero of the polynomial:

q(x) = 4x

Solution 20

q(x) = 0

4x = 0

x = 0

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

Question 21

If 2 and 0 are the zeros of the polynomial f(x) = 2x3
– 5x2 + ax + b then find the values of a and
b.

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

Solution 21

f(x) = 2x3 – 5x2
+ 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

Question 1

By actual division, find the quotient and the remainder
when (x4 + 1) is divided by (x – 1).

Verify that remainder = f(1).

Solution 1

Quotient = x3
+ x2 + x + 1

Remainder = 2

Verification:

f(x) = x4 + 1

Then, f(1) = 14
+ 1 = 1 + 1 = 2 = Remainder

Question 2

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = x3 – 6x2 + 2x – 4, g(x) = .

Solution 2

By
the remainder theorem, we know that when p(x) =
x3 – 6x2 + 2x – 4 is divided by
g(x) = , the remainder is g.

Now,

Hence,
the required remainder is .

Question 3

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = 2x3 + 3x2 – 11x – 3, g(x) = .

Solution 3

By
the remainder theorem, we know that when p(x) =
2x3 + 3x2 – 11x – 3 is divided by
g(x) = , the remainder is g.

Now,

Hence,
the required remainder is 3.

Question 4

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = x3 – ax2 + 6x – a, g(x) = x –
a.

Solution 4

x
– a = 0

x = a

By
the remainder theorem, we know that when p(x) =
x3 – ax2 + 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
= a3– a3+ 6a – a = 5a

Hence,
the required remainder is 5a.

Question 5

The polynomial (2x3 + x2ax + 2) and (2x3 – 3x2 – 3x + a)
when divided by (x – 2) leave the same remainder. Find the value of a.

Solution 5

Let p(x) = 2x3 + x2
ax + 2 and q(x) = 2x3 – 3x2
– 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

Question 6

The polynomial f(x) = x4 – 2x3 + 3x2 – 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).

Solution 6

Letf(x) = (x4 – 2x3 + 3x2 – ax + b)

From the given information,

f(1) = 14 – 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) = x4 – 2x3 + 3x2 – ax + b

= x4 – 2x3 + 3x2 – 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.

Question 7

If p(x) = x3 – 5x2 + 4x – 3 and g(x)
= x – 2, show that p(x) is not a multiple of g(x).

Solution 7

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).

Question 8

If p(x) = 2x3 – 11x2 – 4x + 5 and
g(x) = 2x + 1, show that g(x) is not a factor of p(x).

Solution 8

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).

Question 9

Verify the division algorithm for the polynomials

p(x) = 2x4 – 6x3 + 2x2 – x
+ 2 and g(x) = x + 2.

Solution 9

Question 10

Using remainder theorem, find the remainder when:

(x3 – 6x2 + 9x + 3) is divided by (x – 1)

Solution 10

f(x) = x3 – 6x2 + 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) = 13 – 6 12 + 9 1 + 3

= 1 – 6 + 9 + 3

= 13 – 6 = 7

The required remainder is 7.

Question 11

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = 2x3 – 7x2 + 9x – 13, g(x) = x –
3.

Solution 11

x
– 3 = 0

x = 3

By
the remainder theorem, we know that when p(x) =
2x3 – 7x2 + 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.

Question 12

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = 3x4 – 6x2 – 8x – 2, g(x) = x –
2.

Solution 12

x
– 2 = 0

x = 2

By
the remainder theorem, we know that when p(x) =
3x4 – 6x2 – 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.

Question 13

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = 2x3 – 9x2 + x + 15, g(x) = 2x –
3.

Solution 13

2x
– 3 = 0

x =

By
the remainder theorem, we know that when p(x) =
2x3 – 9x2 + x + 15 is divided by
g(x) = 2x – 3, the remainder is g.

Now,

Hence,
the required remainder is 3.

Question 14

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = x3 – 2x2 – 8x – 1, g(x) = x + 1.

Solution 14

x
+ 1 = 0

x = -1

By
the remainder theorem, we know that when p(x) =
x3 – 2x2 – 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.

Question 15

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = 2x3 + x2– 15x – 12, g(x) = x + 2.

Solution 15

x
+ 2 = 0

x = -2

By
the remainder theorem, we know that when p(x) = 2
x3 + x2 – 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.

Question 16

Using the remainder theorem, find the remainder, when p(x)
is divided by g(x), where

p(x) = 6x3 + 13x2 + 3, g(x) = 3x + 2.

Solution 16

3x
+ 2 = 0

x =

By
the remainder theorem, we know that when p(x) =
6x3 + 13x2 + 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

Question 1

Using factor theorem, show that:

(x – 2) is a factor of (x3 – 8)

Solution 1

f(x) = (x3 – 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 (x3 – 8).

Question 2

Using factor theorem, show that:

(x + ) is a factor of

Solution 2

f(x) =

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

Here,

Question 3

Show that (p – 1) is a factor of (p10 – 1) and
also of (p11 – 1).

Solution 3

Let q(p) = (p10 – 1)
and f(p) = (p11 – 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) = p10 – 1

q(1) = 110 – 1 = 1 – 1 = 0

Hence, (p – 1) is a factor of p10
– 1.

And, f(p) = p11 – 1

f(1) = 111 – 1 = 1 – 1 = 0

Hence, (p – 1) is also a factor of
p11 – 1.

Question 4

Find the value of k for which (x – 1) is a factor of (2x3+ 9x2 + x + k).

Solution 4

f(x) = (2x3 + 9x2 + x + k)

x – 1 = 0 x = 1

f(1) = 2 13 + 9 12 + 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.

Question 5

Find the value of a for which (x – 4) is a factor of (2x3 – 3x2 – 18x + a).

Solution 5

f(x) = (2x3 – 3x2 – 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

Question 6

Find the value of a for which (x
+ 1) is a factor of (ax3 + x2 – 2x + 4a – 9).

Solution 6

Let
p(x) =
ax3 + x2
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

Question 7

Find the value of a for which (x
+ 2a) is a factor of (x5 – 4a2x3 + 2x + 2a +
3).

Solution 7

Let
p(x) =
x5 – 4a2x3
+ 2x + 2a + 3

It is given that (x + 2a) is a
factor of p(x).

p(-2a) = 0

(-2a)5 – 4a2(-2a)3 + 2(-2a) + 2a + 3 = 0

-32a5
– 4a2(-8a3) – 4a + 2a + 3 = 0

-32a5 + 32a5 -2a + 3 = 0

2a = 3

Question 8

Find the value of m for which (2x – 1) is a factor of (8x4
+ 4x3 – 16x2 + 10x + m).

Solution 8

Let
p(x) =
8x4 + 4x3
16x2 + 10x + m

It is given that (2x – 1) is a
factor of p(x).

Question 9

Find the value of a for which the
polynomial (x4 – x3 – 11x2 – x + a) is
divisible by (x + 3).

Solution 9

Let
p(x) =
x4 – x3
11x2 – 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

Question 10

Without
actual division, show that (x3 – 3x2 – 13x + 15) is
exactly divisible by (x2 + 2x – 3).

Solution 10

Let f(x) = x3
– 3x2 – 13x + 15

Now, x2 + 2x – 3 = x2
+ 3x – x – 3

= x (x + 3) – 1
(x + 3)

= (x +
3) (x – 1)

Thus,
f(x) will be exactly divisible by x2 + 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) = 13 – 3 12
13 1 + 15

= 1 – 3 –
13 + 15

= 16 – 16 =
0

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

So,
x2 + 2x – 3 divides f(x) exactly.

Question 11

If (x3 + ax2 + 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.

Solution 11

Letf(x) = (x3 + ax2 + bx + 6)

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

So, f(3) = 33 + a 32 + 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) =  23 + a 22 + 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.

Question 12

Using factor theorem, show that:

(x – 3) is a factor of (2x3 + 7x2 – 24x – 45)

Solution 12

f(x) = (2x3 + 7x2 – 24x – 45)

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

Here, f(3) = 2 33 + 7 32 – 24 3 – 45

= 54 + 63 – 72 – 45

= 117 – 117 = 0

(x – 3) is a factor of (2x3 + 7x2 – 24x – 45).

Question 13

Find the values of a and b so that the polynomial (x3 – 10x2 + ax + b) is exactly divisible by (x – 1) as well as (x – 2).

Solution 13

Let f(x) = (x3 – 10x2 + 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) = 13 – 10 12 + a 1 + b = 0

1 – 10 + a + b = 0

a + b = 9(i)

Andf(2) = 23 – 10 22 + 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.

Question 14

Find the values of a and b so that the polynomial (x4 + ax3 – 7x2 – 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

Solution 14

Letf(x)= (x4 + ax3 – 7x2 – 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.

Question 15

If both (x – 2) and  are factors of px2
+ 5x + r, prove that p = r.

Solution 15

Let
f(x) = px2 + 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

Question 16

Without actual division, prove that 2x4 – 5x3
+ 2x2 – x + 2 is divisible by x2 – 3x + 2.

Solution 16

Let f(x) = 2x4 – 5x3
+ 2x2 – x + 2

and g(x) = x2 – 3x + 2

=
x2 – 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).

Question 17

What must be added to 2x4 – 5x3 + 2x2
– x – 3 so that the result is exactly divisible by (x – 2)?

Solution 17

Let the required number to be

Then, p(x) = 2x4 – 5x3
+ 2x2 – 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

Question 18

What must be subtracted from (x4 + 2x3
– 2x2 + 4x + 6) so that the result is exactly divisible by (x2
+ 2x – 3)?

Solution 18

Let p(x) = x4 + 2x3
– 2x2 + 4x + 6 and q(x) = x2 + 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)

= (x4 + 2x3
– 2x2 + 4x + 6) – (ax + b)

= x4 + 2x3
2x2 + (4 – a)x + 6 – b

We have,

q(x) = x2 + 2x – 3

=
x2 + 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.

Question 19

Use factor theorem to prove that (x + a) is a factor of (xn + an) for any odd positive
integer n.

Solution 19

Let f(x) = xn
+ an

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 + an

=
(-1)n an + an

=
[(-1)n + 1] an

=
[-1 + 1] an …[n is odd

(-1)n = -1]

=
0
×
an

=
0

Hence, (x + a) is a factor of xn + an for any odd positive
integer n.

Question 20

Using factor theorem, show that:

(x – 1) is a factor of (2x4 + 9x3 + 6x2 – 11x – 6)

Solution 20

f(x) = (2x4 + 9x3 + 6x2 – 11x – 6)

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

Here, f(1) = 2 14 + 9 13 + 6 12 – 11 1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17 = 0

(x – 1) is factor of (2x4 + 9x3 + 6x2 – 11x – 6).

Question 21

Using factor theorem, show that:

(x + 2) is a factor of (x4 – x2 – 12)

Solution 21

f(x) = (x4 – x2 – 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 (x4 – x2 – 12).

Question 22

p(x) = 69 + 11x – x2 + x3, g(x) = x
+ 3

Solution 22

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

Now, p(x) = 69 + 11x – x2
+ x3

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).

Question 23

Using factor theorem, show that:

(x + 5) is a factor of (2x3 + 9x2 – 11x – 30)

Solution 23

f(x) = 2x3 + 9x2 – 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 (2x3 + 9x2 – 11x – 30).

Question 24

Using factor theorem, show that:

(2x – 3) is a factor of (2x4 + x3 – 8x2 – x + 6)

Solution 24

f(x) = (2x4 + x3 – 8x2 – 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 .

Question 25

p(x) = 3x3 + x2 – 20x + 12, g(x) =
3x – 2

Solution 25

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

Now, p(x) = 3x3 + x2
– 20x + 12

Hence, g(x) = 3x – 2 is a factor
of the given polynomial p(x).

Question 26

Using factor theorem, show that:

(x – ) is a factor of

Solution 26

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

Question 1

Which of the following
expressions is a polynomial in one variable?

Solution 1

Question 2

Degree of the zero
polynomial is

(a) 1

(b) 0

(c) not defined

(d)none of these

Solution 2

Question 3

Zero of the zero
polynomial is

(a) 0

(b) 1

(c) every real
number

(d)not defined

Solution 3

Question 4

If p(x) = x + 4, then
p(x) + p(-x) = ?

(a) 0

(b) 4

(c) 2x

(d)8

Solution 4

Question 5

Solution 5

Question 6

If p(x) = 5x – 4x2 + 3 then p(-1) = ?

(a) 2

(b) -2

(c) 6

(d) -6

Solution 6

Correct
option: (d)

P(x)
= 5x – 4x2 + 3

p(-1) = 5(-1) – 4(-1)2
+ 3 = -5 – 4 + 3 = -6

Question 7

If (x51 + 51) is divided by (x + 1) then the
remainder is

(a) 0

(b) 1

(c) 49

(d) 50

Solution 7

Correct
option: (d)

Let
f(x) = x51 + 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

Question 8

If (x + 1) is a factor of the polynomial (2x2 + kx) then k = ?

(a)   4

(b)   -3

(c)  2

(d) -2

Solution 8

Correct option: (c)

Let p(x) = 2x2 + kx

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

P(–1) = 0

⇒ 2(–1)2 + k(–1) = 0

⇒ 2 – k = 0

⇒ k = 2

Question 9

When p(x) = x4
+ 2x3 – 3x2 + x – 1 is divided by (x – 2), the
remainder is

(a) 0

(b) -1

(c) -15

(d)21

Solution 9

Question 10

When p(x) = x3
– 3x+ 4x + 32 is divided
by (x + 2), the remainder is

(a) 0

(b) 32

(c) 36

(d)4

Solution 10

Question 11

When p(x) = 4x3
– 12x2 + 11x – 5 is divided by (2x – 1), the remainder is

(a) 0

(b) -5

(c) -2

(d)2

Solution 11

Question 12

Which of the following
expressions is a polynomial?

Solution 12

Question 13

When p(x) =x3-ax2+x
is divided by (x-a), the remainder is

(a) 0

(b) a

(c) 2a

(d)3a

Solution 13

Question 14

When p(x) = x3
+ ax2 + 2x + a is divided by (x + a), the remainder is

(a) 0

(b) a

(c) -a

(d)2a

Solution 14

Question 15

(x + 1) is a factor of
the polynomial

(a) x3
2x2 + x + 2

(b) x3 +
2x2 + x – 2

(c) x3 +
2x2 – x – 2

(d)x3 +
2x2 – x + 2

Solution 15

Question 16

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

(a)

(b)

(c)

(d)

Solution 16

Correct
option: (b)

p(x)
= 2x + 5

Now,
p(x) = 0

2x + 5 = 0

2x = -5

Question 17

The zeroes of the
polynomial p(x) = x2 + x – 6 are

(a) 2, 3

(b) -2, 3

(c) 2, -3

(d)-2, -3

Solution 17

Question 18

The zeroes of the
polynomial p(x) = 2x2 + 5x – 3 are

Solution 18

Question 19

The zeros of the polynomial p(x) = 2x2 + 7x – 4
are

(a)

(b)

(c)

(d)

Solution 19

Correct
option: (c)

p(x)
= 2x2 + 7x – 4

Now,
p(x) = 0

2x2 + 7x – 4 = 0

2x2 + 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 =

Question 20

If (x + 5) is a factor
of p(x) = x3 – 20x + 5k, then k =?

(a) -5

(b) 5

(c) 3

(d)-3

Solution 20

Question 21

If (x + 2) and (x – 1)
are factors of (x3 + 10x2 + mx
+ n), then

(a) m = 5, n = -3

(b) m = 7, n = -18

(c) m = 17, n = -8

(d)m = 23, n = -19

Solution 21

Question 22

If (x100 + 2x99
+ k) is divisible by (x + 1), then the value of k is

(a) 1

(b) 2

(c) -2

(d)-3

Solution 22

Question 23

Which of the following
is a polynomial?

Solution 23

Question 24

For what value of k is
the polynomial p(x) = 2x3 – kx2 + 3x + 10 exactly
divisible by (x + 2)?

Solution 24

Question 25

The zeroes of the polynomial
p(x) = x2 – 3x are

(a) 0, 0

(b) 0, 3

(c) 0, -3

(d)3, -3

Solution 25

Question 26

The zeros of the polynomial p(x) = 3x2 – 1 are

(a)

(b)

(c)

(d)

Solution 26

Correct
option: (d)

p(x)
= 3x2 – 1

Now,
p(x) = 0

3x2
1 = 0

3x2
= 1

Question 27

Which of the following
is a polynomial?

Solution 27

Question 28

Which of the following
is a polynomial?

(a) x-2
+ x-1 + 3

(b) x + x-1
+ 2

(c) x-1

(d)0

Solution 28

Question 29

Which of the following

(a) x + 4

(b) x3 +
x

(c) x3 +
2x + 6

(d)x2 +
5x + 4

Solution 29

Question 30

Which of the following
is a linear polynomial?

(a) x + x2

(b) x + 1

(c) 5x2
– x + 3

(d)

Solution 30

Question 31

Which of the following
is a binomial?

(a) x2 + x + 3

(b) x2 + 4

(c) 2x2

(d)

Solution 31

Question 32

(a)

(b) 2

(c) 1

(d)0

Solution 32

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