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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials

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Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 2 - Polynomials.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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Exercise/Page

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

  

Solution 12

  

Solution 13

Solution 14

Correct option: (d)

P(x) = 5x - 4x2 + 3

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

Solution 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 = 50

Solution 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 = 2

Solution 17

  

Solution 18

  

Solution 19

  

Solution 20

  

Solution 21

  

Solution 22

Solution 23

Correct option: (b)

p(x) = 2x + 5

Now, p(x) = 0

2x + 5 = 0

2x = -5

Solution 24

Solution 25

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 =   

Solution 27

Solution 28

  

Solution 29

  

Solution 30

  

Solution 31

Solution 32

Correct option: (d)

p(x) = 3x2 - 1

Now, p(x) = 0

3x2 - 1 = 0

3x2 = 1

  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2A

Solution 1(v)

It is a polynomial, Degree = 2.

Solution 1(vi)

It is not a polynomial.

Solution 1(vii)

It is a polynomial, Degree = 0.

Solution 1(viii)

It is a polynomial, Degree = 0.

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. 

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. 

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. 

Solution 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. 

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.

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. 

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. 

Solution 1(xii)

  

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

So, it is not a polynomial. 

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. 

Solution 1(xiv)

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. 

Solution 1(xv)

  

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. 

Solution 2(i)

-7 + x

The degree of a given polynomial is 1.

Hence, it is a linear polynomial.

Solution 2(ii)

6y

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. 

Solution 2(iii)

-z3

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. 

Solution 2(iv)

1 - y - y3

The degree of a given polynomial is 3.

Hence, it is a cubic polynomial. 

Solution 2(v)

x - x3 + x4

The degree of a given polynomial is 4.

Hence, it is a quartic polynomial. 

Solution 2(vi)

1 + x + x2

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. 

Solution 2(vii)

-6x2

The degree of a given polynomial is 2.

Hence, it is a quadratic polynomial. 

Solution 2(viii)

-13

The given polynomial contains only one term namely constant.

Hence, it is a constant polynomial. 

Solution 2(ix)

-p

The degree of a given polynomial is 1.

Hence, it is a linear polynomial. 

Solution 3(i)

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

Solution 3(ii)

The coefficient of x in  . 

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. 

Solution 3(iv)

The coefficient of x in  . 

Solution 3(v)

The constant term in  . 

Solution 4(i)

  

Hence, the degree of a given polynomial is 2. 

Solution 4(ii)

y2(y - y3)

= y3 - y5

Hence, the degree of a given polynomial is 5. 

Solution 4(iii)

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

= 6x4 + 9x3 - 4x3 - 6x2

= 6x4 + 5x3 - 6x2

Hence, the degree of a given polynomial is 4. 

Solution 4(iv)

  

The degree of a given polynomial is 1. 

Solution 4(v)

-8

This is a constant polynomial.

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

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. 

Solution 5(i)

Example of a monomial of degree 5:

3x5 

Solution 5(ii)

Example of a binomial of degree 8:

x - 6x8 

Solution 5(iii)

Example of a trinomial of degree 4:

7 + 2y + y4 

Solution 5(iv)

Example of a monomial of degree 0:

7 

Solution 6(i)

x - 2x2 + 8 + 5x3 in standard form:

5x3 - 2x2 + x + 8 

Solution 6(ii)

 

   

Solution 6(iii)

6x3 + 2x - x5 - 3x2 in standard form:

-x5 + 6x3 - 3x2 + 2x 

Solution 6(iv)

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

t4 - 3t3 - t2 + t + 2 

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2B

Solution 1

p(x) = 5 - 4x + 2x2

(i) p(0) = 5 - 4 0 + 2 02 = 5

(ii) p(3) = 5 - 4 3 + 2 32

= 5 - 12 + 18

= 23 - 12 = 11

(iii) p(-2) = 5 - 4(-2) + 2 (-2)2

= 5 + 8 + 8 = 21

Solution 2

p(y) = 4 + 3y - y2 + 5y3

(i) p(0) = 4 + 3 0 - 02 + 5 03

= 4 + 0 - 0 + 0 = 4

(ii) p(2) = 4 + 3 2 - 22 + 5 23

= 4 + 6 - 4 + 40

= 10 - 4 + 40 = 46

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

= 4 - 3 - 1 - 5 = -5

Solution 3

f(t) = 4t2 - 3t + 6

(i) f(0) = 4 02 - 3 0 + 6

= 0 - 0 + 6 = 6

(ii) f(4) = 4(4)2 - 3 4 + 6

= 64 - 12 + 6 = 58

(iii) f(-5) = 4(-5)2 - 3(-5) + 6

= 100 + 15 + 6 = 121

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.  

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. 

Solution 6(i)

p(x) = x - 4

Then, p(4) = 4 - 4 = 0

 4 is a zero of the polynomial p(x).

Solution 6(ii)

p(x) = x - 3

Then,p(-3) = -3 - 3 = -6

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

Solution 6(iii)

p(y) = 2y + 1

Then,

is a zero of the polynomial p(y).

Solution 6(iv)

p(x) = 2 - 5x

Then, 

 is a zero of the polynomial p(x).

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).

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).

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).

Solution 8(i)

p(x) = 0

x - 5 = 0

x = 5

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

Solution 8(ii)

q(x) = 0

 x + 4 = 0

x= -4

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

Solution 8(iv)

f(x) = 0

 3x + 1= 0

3x=-1

 x =

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

Solution 8(v)

g(x) = 0

 5 - 4x = 0

 -4x = -5

 x =

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

Solution 8(vii)

p(x) = 0

 ax = 0

x = 0

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

Solution 8(viii)

q(x) = 0

4x = 0

 x = 0

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

Solution 8(iii)

r(x) = 2x + 5

Now, r(x) = 0

2x + 5 = 0

2x = -5

  

Solution 8(vi)

h(x) = 6x - 2

Now, h(x) = 0

6x - 2 = 0

6x = 2

  

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. 

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2C

Solution 1

Quotient = x3 + x2 + x + 1

Remainder = 2

Verification:

f(x) = x4 + 1

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

Solution 2

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.

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.

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. 

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. 

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. 

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. 

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. 

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  . 

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. 

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. 

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 

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.

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).

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). 

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 2 - Polynomials Page/Excercise 2D

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).


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).

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).

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).

Solution 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). 

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).

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 .

Solution 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). 

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

Solution 10

f(x) =

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

Here, 

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. 

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.


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


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 

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

Solution 16

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

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

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 

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.

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.

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.

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.

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 = r

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). 

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. 

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. 

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. 

TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 2 - Polynomials for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 2 - Polynomials now.

Text Book Solutions

CBSE IX - Mathematics

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