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R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 9 Chapter 2 Polynomial

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 1

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 2

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

The given expression can be written as R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

Question 6

Which of the expressions are polynomials?

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 6

It is a polynomial, Degree = 2.

Question 7

Which of the expressions are polynomials?

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - 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?

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 11

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 13

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 15

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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,
it is a quadratic polynomial.

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,
it is a quadratic polynomial.

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Solution 26

The coefficient of x in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Solution 28

The coefficient of x in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Question 29

Write the constant term in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Solution 29

The constant term in R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Question 30

Determine the degree of each of the following polynomials.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 30

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 33

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 41

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 0 + 2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 02 = 5

(ii) p(3) = 5 – 4 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 3 + 2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 0 – 02 + 5 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 03

= 4 + 0 – 0 + 0 = 4

(ii) p(2) = 4 + 3 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 2 – 22 + 5 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 02 – 3 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 0 + 6

= 0 – 0 + 6 = 6

(ii) f(4) = 4(4)2 – 3 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials4 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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials -3 is not a zero of the polynomial p(x).

Question 8

Verify that:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsis a zero of the polynomial p(y) = 2y + 1.

Solution 8

p(y) = 2y + 1

Then, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsis a zero of the polynomial p(y).

Question 9

Verify that:

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsis a zero of thepolynomial p(x) = 2 – 5x.

Solution 9

p(x) = 2 – 5x

Then, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsis 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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-1 = 0

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials1 is a zero of the polynomial p(x).

Also, p(2) = (2 – 1)(2 – 2) = 1 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials0 = 0

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials2 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

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials2 is a zero of the polynomial p(x).

 Also, p(-3) = (-3)2 – 3 – 6

= 9 – 3 – 6 = 0

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials0 = 0

p(3) = (3)2– 3 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials3 = 9 – 9 = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials0 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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsx – 5 = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsx = 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials5 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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x + 4 = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsx= -4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials -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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

Question 16

Find the zero of the polynomial:

f(x) = 3x + 1

Solution 16

f(x) = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 3x + 1= 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials3x=-1

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x =R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x =R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsis the zero of the polynomial f(x).

Question 17

Find the zero of the polynomial:

g(x) = 5 – 4x

Solution 17

g(x) = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 5 – 4x = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials -4x = -5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x =R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 19

Find the zero of the polynomial:

p(x) = ax, a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 0

Solution 19

p(x) = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials ax = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsx = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials0 is the zero of the polynomial p(x).

Question 20

Find the zero of the polynomial:

q(x) = 4x

Solution 20

q(x) = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials4x = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials0 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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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) = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Solution 2

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

By
the remainder theorem, we know that when p(x) =
x3 – 6x2 + 2x – 4 is divided by
g(x) = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials, the remainder is gR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Now,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Hence,
the required remainder is R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

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) = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Solution 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 

By
the remainder theorem, we know that when p(x) =
2x3 + 3x2 – 11x – 3 is divided by
g(x) = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials, the remainder is gR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Now,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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)

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - PolynomialsFrom the given information,

f(1) = 14 – 2(1)3 + 3(1)2 – a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials1 + b = 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 1 – 2 + 3 – a + b = 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 2 – a + b = 5(i)

And,

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 1 + 2 + 3 + a + b = 19

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 6 + a + b = 19(ii)

Adding (i) and (ii), we get

8 + 2b = 24

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials2b= 24 – 8 = 16

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsb = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

2 – a + 8 = 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-a + 10 = 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-a = -10 + 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-a = -5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsa = 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsa = 5 and b = 8

f(x) = x4 – 2x3 + 3x2 – ax + b

= x4 – 2x3 + 3x2 – 5x + 8

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsf(2) = (2)4 – 2(2)3 + 3(2)2 – 5 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials2 + 8

= 16 – 16 + 12 – 10 + 8

= 20 – 10 = 10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - PolynomialsThe 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 = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Also,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 12 + 9 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 1 + 3

= 1 – 6 + 9 + 3

= 13 – 6 = 7

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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 = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

By
the remainder theorem, we know that when p(x) =
2x3 – 9x2 + x + 15 is divided by
g(x) = 2x – 3, the remainder is gR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Now,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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 = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

By
the remainder theorem, we know that when p(x) =
6x3 + 13x2 + 3 is divided by
g(x) = 3x + 2, the remainder is gR-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

Now,

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials (x – 2) is a factor of (x3 – 8).

 

Question 2

Using factor theorem, show that:

(x + R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials) is a factor of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 2

f(x) = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

Here, 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x = 1

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials f(1) = 2 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 13 + 9 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsf(1) = 12 + k = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsk = -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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x = 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials f(4) = 2(4)3 – 3(4)2 – 18 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials f(4) = 8 + a = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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).

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials32 + b R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials3 + 6 = 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials27 + 9a + 3b + 6 = 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials9 a + 3b + 33 = 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials9a + 3b = 3 – 33

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials9a + 3b = -30

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials3a + 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.

 

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials f(2) =  23 + a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials22 + b R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials2 + 6 = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials                       8 + 4a+ 2b + 6 = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials                               4a + 2b = -14

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials                                   2a + b = -7(ii)

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsa = -3

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

3(-3) + b = -10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-9 + b = -10

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsb = -10 + 9

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsb = -1

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsa = -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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 33 + 7 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 32 – 24 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 3 – 45

= 54 + 63 – 72 – 45

= 117 – 117 = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials (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.

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsf(1) = 13 – 10 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials12 + a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials1 + b = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials1 – 10 + a + b = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsa + b = 9(i)

Andf(2) = 23 – 10 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials22 + a R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials2 + b = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials8 – 40 + 2a + b = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials2a + b = 32(ii)

Subtracting (i) from (ii), we get

a = 23

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

23 + b = 9

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsb = 9 – 23

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsb = -14

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsa = 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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsx = -2 and x + 3 = 0 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsx = -3

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsf(-2) = (-2)4 + a (-2)3 – 7 (-2)2 – 8 (-2) + b = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials16 – 8a – 28 + 16 + b = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-8a + b = -4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials8a – b = 4(i)

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials81 – 27a – 63 + 24 + b = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-27a + b = -42

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials27a – 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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials2 – b = 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials16 – b = 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-b = -16 + 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials-b = -12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsb = 12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomialsa = 2 and b = 12.

Question 15

If both (x – 2) and R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 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,
R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials is a factor of
f(x).

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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
added be k.

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
added is 5.

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 14 + 9 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 13 + 6 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 12 – 11 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials 1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17 = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials (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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials (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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials (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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials x = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials is a factor of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials.

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 R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials
 = 0.

Now, p(x) = 3x3 + x2
– 20x + 12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

Question 26

Using factor theorem, show that:

(x – R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials) is a factor of R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 26

f(x) = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

Here, R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

= 14 – 8 – 6

= 14 – 14 = 0

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Chapter 2 – Polynomials Exercise MCQ

Question 1

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 1

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 2

Degree of the zero
polynomial is

(a) 1

(b) 0

(c) not defined

(d)none of these

Solution 2

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 3

Zero of the zero
polynomial is

(a) 0

(b) 1

(c) every real
number

(d)not defined

Solution 3

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 4

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

(a) 0

(b) 4

(c) 2x

(d)8

Solution 4

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 12

Which of the following
expressions is a polynomial?

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 12

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 16

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

(a) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(b) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(c) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(d) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 16

Correct
option: (b)

p(x)
= 2x + 5

Now,
p(x) = 0

2x + 5 = 0

2x = -5

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 18

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 18

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 19

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

(a) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(b) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(c) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(d) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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 = R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 23

Which of the following
is a polynomial?

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 23

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 24

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 24

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 26

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

(a) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(b) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(c) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(d) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 26

Correct
option: (d)

p(x)
= 3x2 – 1

Now,
p(x) = 0

3x2
1 = 0

3x2
= 1

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 27

Which of the following
is a polynomial?

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 27

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

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

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 29

Which of the following
is a quadratic polynomial?

(a) x + 4

(b) x3 +
x

(c) x3 +
2x + 6

(d)x2 +
5x + 4

Solution 29

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 30

Which of the following
is a linear polynomial?

(a) x + x2

(b) x + 1

(c) 5x2
– x + 3

(d)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 30

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 31

Which of the following
is a binomial?

(a) x2 + x + 3

(b) x2 + 4

(c) 2x2

(d)R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Solution 31

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

Question 32

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(a) R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

(b) 2

(c) 1

(d)0

Solution 32

R-s-aggarwal-and-v-aggarwal Solutions Cbse Class 9 Mathematics Chapter - Polynomials

 

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