RS Agarwal Solution | Class 9th | Chapter-2 | Polynomials | Edugrown

Exercise MCQ

Question 1

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

Solution 1

Question 2

Which of the following expressions is a polynomial?

Solution 2

Question 3

Which of the following is a polynomial?

Solution 3

Question 4

Which of the following is a polynomial?

Solution 4

Question 5

Which of the following is a polynomial?

(a) x^-2 + x^-1 + 3

(b) x + x^-1 + 2

(c) x^-1

(d)0Solution 5

Question 6

Which of the following is a quadratic polynomial?

(a) x + 4

(b) x³ + x

(c) x³ + 2x + 6

(d)x² + 5x + 4Solution 6

Question 7

Which of the following is a linear polynomial?

(a) x + x²

(b) x + 1

(c) 5x² – x + 3

(d)Solution 7

Question 8

Which of the following is a binomial?

(a) x² + x + 3

(b) x² + 4

(c) 2x²

(d)Solution 8

Question 9

(a) 

(b) 2

(c) 1

(d)0Solution 9

Question 10

Degree of the zero polynomial is

(a) 1

(b) 0

(c) not defined

(d)none of theseSolution 10

Question 11

Zero of the zero polynomial is

(a) 0

(b) 1

(c) every real number

(d)not definedSolution 11

Question 12

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

(a) 0

(b) 4

(c) 2x

(d)8Solution 12

Question 13

Solution 13

Question 14

If p(x) = 5x – 4x² + 3 then p(-1) = ?

(a) 2

(b) -2

(c) 6

(d) -6Solution 14

Correct option: (d)

P(x) = 5x – 4x² + 3

⇒ p(-1) = 5(-1) – 4(-1)² + 3 = -5 – 4 + 3 = -6Question 15

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

(a) 0

(b) 1

(c) 49

(d) 50Solution 15

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 = 50Question 16

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

(a)   4

(b)   -3

(c)  2

(d) -2Solution 16

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 = 2Question 17

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

(a) 0

(b) -1

(c) -15

(d)21Solution 17

Question 18

When p(x) = x³ – 3x² + 4x + 32 is divided by (x + 2), the remainder is

(a) 0

(b) 32

(c) 36

(d)4Solution 18

Question 19

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

(a) 0

(b) -5

(c) -2

(d)2Solution 19

Question 20

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

(a) 0

(b) a

(c) 2a

(d)3aSolution 20

Question 21

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

(a) 0

(b) a

(c) -a

(d)2aSolution 21

Question 22

(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 + 2Solution 22

Question 23

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

(a) 

(b) 

(c) 

(d) Solution 23

Correct option: (b)

p(x) = 2x + 5

Now, p(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

Question 24

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

(a) 2, 3

(b) -2, 3

(c) 2, -3

(d)-2, -3Solution 24

Question 25

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

Solution 25

Question 26

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

(a) 

(b) 

(c) 

(d) Solution 26

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 27

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

(a) -5

(b) 5

(c) 3

(d)-3Solution 27

Question 28

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 = -19Solution 28

Question 29

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

(a) 1

(b) 2

(c) -2

(d)-3Solution 29

Question 30

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

Solution 30

Question 31

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

(a) 0, 0

(b) 0, 3

(c) 0, -3

(d)3, -3Solution 31

Question 32

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

(a) 

(b) 

(c) 

(d) Solution 32

Correct option: (d)

p(x) = 3x² – 1

Now, p(x) = 0

⇒ 3x² – 1 = 0

⇒ 3x² = 1

Exercise Ex. 2A

Question 1(v)

Which of the expressions are polynomials?

Solution 1(v)

It is a polynomial, Degree = 2.Question 1(vi)

Which of the expressions are polynomials?

Solution 1(vi)

It is not a polynomial.Question 1(vii)

Which of the expressions are polynomials?

1Solution 1(vii)

It is a polynomial, Degree = 0.Question 1(viii)

Which of the expressions are polynomials?

Solution 1(viii)

It is a polynomial, Degree = 0.Question 1(i)

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

Solution 1(i)

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 1(ii)

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

Solution 1(ii)

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 1(iii)

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

Solution 1(iii)

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 1(iv)

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

X¹⁰⁰ – 1Solution 1(iv)

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 1(ix)

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

Solution 1(ix)

The given expression can be written as  

It contains a term having negative integral power of x. So, it is not a polynomial.Question 1(x)

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

Solution 1(x)

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 1(xi)

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

Solution 1(xi)

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 1(xii)

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

Solution 1(xii)

The given expression contains a term containing x^1/2, where ½ is not a non-negative integer.

So, it is not a polynomial. Question 1(xiii)

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

Solution 1(xiii)

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 1(xiv)

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

x⁴ – x^3/2 + x – 3Solution 1(xiv)

x⁴ – 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. Question 1(xv)

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

Solution 1(xv)

The given expression can be written as 2x³ + 3x² + 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. Question 2(i)

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

-7 + xSolution 2(i)

-7 + x

The degree of a given polynomial is 1.

Hence, it is a linear polynomial.Question 2(ii)

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

6ySolution 2(ii)

6y

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. Question 2(iii)

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

-z³Solution 2(iii)

-z³

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. Question 2(iv)

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

1 – y – y³Solution 2(iv)

1 – y – y³

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. Question 2(v)

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

x – x³ + x⁴Solution 2(v)

x – x³ + x⁴

The degree of a given polynomial is 4.

Hence, it is a quartic polynomial. Question 2(vi)

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

1 + x + x²Solution 2(vi)

1 + x + x²

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. Question 2(vii)

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

-6x²Solution 2(vii)

-6x²

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. Question 2(viii)

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

-13Solution 2(viii)

-13

The given polynomial contains only one term namely constant.

Hence, it is a constant polynomial. Question 2(ix)

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

-pSolution 2(ix)

-p

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. Question 3(i)

Write the coefficient of x³ in x + 3x² – 5x³ + x⁴Solution 3(i)

The coefficient of x³ in x + 3x² – 5x³ + x⁴ is -5. Question 3(ii)

Write the coefficient of x in .Solution 3(ii)

The coefficient of x in . Question 3(iii)

Write the coefficient of x² in 2x – 3 + x³.Solution 3(iii)

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

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

Write the coefficient of x in .Solution 3(iv)

The coefficient of x in . Question 3(v)

Write the constant term in .Solution 3(v)

The constant term in . Question 4(i)

Determine the degree of each of the following polynomials.

Solution 4(i)

Hence, the degree of a given polynomial is 2. Question 4(ii)

Determine the degree of each of the following polynomials.

y²(y – y³)Solution 4(ii)

y²(y – y³)

= y³ – y⁵

Hence, the degree of a given polynomial is 5. Question 4(iii)

Determine the degree of each of the following polynomials.

(3x – 2)(2x³ + 3x²)Solution 4(iii)

(3x – 2)(2x³ + 3x²)

= 6x⁴ + 9x³ – 4x³ – 6x²

= 6x⁴ + 5x³ – 6x²

Hence, the degree of a given polynomial is 4. Question 4(iv)

Determine the degree of each of the following polynomials.

Solution 4(iv)

The degree of a given polynomial is 1. Question 4(v)

Determine the degree of each of the following polynomials.

-8Solution 4(v)

-8

This is a constant polynomial.

The degree of a non-zero constant polynomial is zero. Question 4(vi)

Determine the degree of each of the following polynomials.

x^-2(x⁴ + x²)Solution 4(vi)

x^-2(x⁴ + x²)

= x^-2.x²(x² + 1)

= x⁰ (x² + 1)

= x² + 1

Hence, the degree of a given polynomial is 2. Question 5(i)

Give an example of a monomial of degree 5.Solution 5(i)

Example of a monomial of degree 5:

3x⁵ Question 5(ii)

Give an example of a binomial of degree 8.Solution 5(ii)

Example of a binomial of degree 8:

x – 6x⁸ Question 5(iii)

Give an example of a trinomial of degree 4.Solution 5(iii)

Example of a trinomial of degree 4:

7 + 2y + y⁴ Question 5(iv)

Give an example of a monomial of degree 0.Solution 5(iv)

Example of a monomial of degree 0:

7 Question 6(i)

Rewrite each of the following polynomials in standard form.

x – 2x² + 8 + 5x³Solution 6(i)

x – 2x² + 8 + 5x³ in standard form:

5x³ – 2x² + x + 8 Question 6(ii)

Rewrite each of the following polynomials in standard form.

Solution 6(ii)

Question 6(iii)

Rewrite each of the following polynomials in standard form.

6x³ + 2x – x⁵ – 3x²Solution 6(iii)

6x³ + 2x – x⁵ – 3x² in standard form:

-x⁵ + 6x³ – 3x² + 2x Question 6(iv)

Rewrite each of the following polynomials in standard form.

2 + t – 3t³ + t⁴ – t²Solution 6(iv)

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

t⁴ – 3t³ – t² + t + 2 

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 = 21Question 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 = -5Question 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 = 121Question 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(i)

Verify that:

4 is a zero of the polynomial p(x) = x – 4.Solution 6(i)

p(x) = x – 4

Then, p(4) = 4 – 4 = 0

 4 is a zero of the polynomial p(x).Question 6(ii)

Verify that:

-3 is a zero of the polynomial p(x) = x – 3.Solution 6(ii)

p(x) = x – 3

Then,p(-3) = -3 – 3 = -6

 -3 is not a zero of the polynomial p(x).Question 6(iii)

Verify that:

is a zero of the polynomial p(y) = 2y + 1.Solution 6(iii)

p(y) = 2y + 1

Then, 

is a zero of the polynomial p(y).Question 6(iv)

Verify that:

is a zero of thepolynomial p(x) = 2 – 5x.Solution 6(iv)

p(x) = 2 – 5x

Then, 

 is a zero of the polynomial p(x).Question 7(i)

Verify that:

1 and 2 are the zeros of the polynomial p(x) = (x – 1) (x – 2).Solution 7(i)

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 7(ii)

Verify that:

2 and -3 are the zeros of the polynomial p(x) = x² + x – 6.Solution 7(ii)

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 7(iii)

Verify that:

0 and 3 are the zeros of the polynomial p(x) = x² – 3x.Solution 7(iii)

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 8(i)

Find the zero of the polynomial:

p(x) = x – 5Solution 8(i)

p(x) = 0

x – 5 = 0

x = 5

_{5 is the zero of the polynomial p(x).}Question 8(ii)

Find the zero of the polynomial:

q(x) = x + 4Solution 8(ii)

q(x) = 0

 x + 4 = 0

x= -4

 _{-4 is the zero of the polynomial q(x).} Question 8(iv)

Find the zero of the polynomial:

f(x) = 3x + 1Solution 8(iv)

f(x) = 0

 3x + 1= 0

3x=-1

 x =

  x =is the zero of the polynomial f(x).Question 8(v)

Find the zero of the polynomial:

g(x) = 5 – 4xSolution 8(v)

g(x) = 0

 5 – 4x = 0

 -4x = -5

 x =

 x =  is the zero of the polynomial g(x).Question 8(vii)

Find the zero of the polynomial:

p(x) = ax, a  0Solution 8(vii)

p(x) = 0

 ax = 0

x = 0

0 is the zero of the polynomial p(x).Question 8(viii)

Find the zero of the polynomial:

q(x) = 4xSolution 8(viii)

q(x) = 0

4x = 0

 x = 0

0 is the zero of the polynomial q(x).Question 8(iii)

Find the zero of the polynomial:

r(x) = 2x + 5Solution 8(iii)

r(x) = 2x + 5

Now, r(x) = 0

⇒ 2x + 5 = 0

⇒ 2x = -5

Question 8(vi)

Find the zero of the polynomial:

h(x) = 6x – 2Solution 8(vi)

h(x) = 6x – 2

Now, h(x) = 0

⇒ 6x – 2 = 0

⇒ 6x = 2

Question 9

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 9

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. 

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

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 3

Using factor theorem, show that:

(x – 1) is a factor of (2x⁴ + 9x³ + 6x² – 11x – 6)Solution 3

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 4

Using factor theorem, show that:

(x + 2) is a factor of (x⁴ – x² – 12)Solution 4

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 5

p(x) = 69 + 11x – x² + x³, g(x) = x + 3Solution 5

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 6

Using factor theorem, show that:

(x + 5) is a factor of (2x³ + 9x² – 11x – 30)Solution 6

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 7

Using factor theorem, show that:

(2x – 3) is a factor of (2x⁴ + x³ – 8x² – x + 6)Solution 7

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 8

p(x) = 3x³ + x² – 20x + 12, g(x) = 3x – 2Solution 8

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 9

Using factor theorem, show that:

(x – ) is a factor of Solution 9

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

Question 10

Using factor theorem, show that:

(x + ) is a factor of Solution 10

f(x) = 

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

Here, 

Question 11

Show that (p – 1) is a factor of (p¹⁰ – 1) and also of (p¹¹ – 1).Solution 11

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 12

Find the value of k for which (x – 1) is a factor of (2x³+ 9x² + x + k).Solution 12

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 13

Find the value of a for which (x – 4) is a factor of (2x³ – 3x² – 18x + a).Solution 13

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 14

Find the value of a for which (x + 1) is a factor of (ax³ + x² – 2x + 4a – 9).Solution 14

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 15

Find the value of a for which (x + 2a) is a factor of (x⁵ – 4a²x³ + 2x + 2a + 3).Solution 15

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 16

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

Let p(x) = 8x⁴ + 4x³ – 16x² + 10x + m

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

Question 17

Find the value of a for which the polynomial (x⁴ – x³ – 11x² – x + a) is divisible by (x + 3).Solution 17

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 18

Without actual division, show that (x³ – 3x² – 13x + 15) is exactly divisible by (x² + 2x – 3).Solution 18

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 19

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 19

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 20

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 20

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 21

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 21

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 22

If both (x – 2) and  are factors of px² + 5x + r, prove that p = r.Solution 22

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 = rQuestion 23

Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.Solution 23

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 24

What must be added to 2x⁴ – 5x³ + 2x² – x – 3 so that the result is exactly divisible by (x – 2)?Solution 24

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 25

What must be subtracted from (x⁴ + 2x³ – 2x² + 4x + 6) so that the result is exactly divisible by (x² + 2x – 3)?Solution 25

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 26

Use factor theorem to prove that (x + a) is a factor of (xⁿ + aⁿ) for any odd positive integer n.Solution 26

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. 

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

Verify the division algorithm for the polynomials

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

Question 3

Using remainder theorem, find the remainder when:

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

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 4

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 4

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 5

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 5

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 6

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 6

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 7

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 7

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 8

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 8

x + 2 = 0

⇒ x = -2

By the remainder theorem, we know that when p(x) = 2x³ + 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 9

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 9

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. Question 10

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 10

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 11

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 11

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 12

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 12

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 13

The polynomial (2x³ + x² – ax + 2) and (2x³ – 3x² – 3x + a) when divided by (x – 2) leave the same remainder. Find the value of a.Solution 13

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 14

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 14

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 15

If p(x) = x³ – 5x² + 4x – 3 and g(x) = x – 2, show that p(x) is not a multiple of g(x).Solution 15

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 16

If p(x) = 2x³ – 11x² – 4x + 5 and g(x) = 2x + 1, show that g(x) is not a factor of p(x).Solution 16

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).