# RS Aggarwal Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2B

R S Aggarwal Solutions for Class 10 Maths Chapter 2 Polynomials, contains solutions for all exercise 2B questions. Chapter 2 RS Aggarwal solutions on exercise 2b is based on polynomials and relation between the zeros and coefficients of a quadratic Polynomial. Students can download the R S Aggarwal Solutions of Class 10 and start practicing offline.

### Access other exercise solutions of Class 10 Maths Chapter 2 Polynomials

Exercise 2A Solutions: 21 Questions (Short Answers)

Exercise 2C Solutions: 25 Questions (Short Answers)

## Exercise 2B Page No: 58

Question 1: Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x^3 â€“ 2x^2 â€“ 5x + 6) and verify the relation between it zeros and coefficients.

Solution:

Let f(x) = x^3 â€“ 2x^2 â€“ 5x + 6

3, -2 and 1 are the zeros of the polynomial (given)

Therefore,

f(3) = (3)^3 â€“ 2(3)^2 â€“ 5(3) + 6

= 27 â€“ 18 â€“ 15 + 6

= 0

f(-2) = (-2)^3 â€“ 2(-2)^2 â€“ 5(-2) + 6

= -8 â€“ 8 + 10 + 6

= 0

f(1) = (1)^3 â€“ 2 (1)^2 â€“ 5 (1) + 6

= 1 â€“ 2 â€“ 5 + 6

= 0

Verify relations:

General form of a cubic equation: ax^3 + bx^2 + cx + d.

Now,

Consider Î± = 3, Î² = – 2 and Î³ = 1

Î± + Î² + Î³ = 3 â€“ 2 + 1 = 2 = -b/a

Î±Î² + Î²Î³ + Î±Î³ = 3 (-2) + (-2) (1) + 1 (3) = -5 = c/a

And, Î±Î²Î³ = 3 (-2)(1) = -6 = -d/a

Question 2: Verify that 5, -2 and 1/3 are the zeroes of the cubic polynomial p(x) = (3x^3 â€“ 10x^2 â€“ 27x + 10) and verify the relation between its zeroes and coefficients.

Solution:

Let f(x) = 3x^3 â€“ 10x^2 â€“ 27x + 10

5, -2 and 1/3 are the zeros of the polynomial (given)

Therefore,

f(5) = 3(5)^3 â€“ 10(5)^2 â€“ 27(5) + 10

= 3 Ã— 125 â€“ 250 â€“ 135 + 10

= 0

f(-2) = 3(-2)^3 â€“ 10(-2)^2 â€“ 27(-2) + 10

= – 24 â€“ 40 + 54 + 10

= 0

f(1/3) = 3 (1/3)^3 â€“ 10(1/3)^2 â€“ 27(1/3) + 10

= 1/9 â€“ 10/9 â€“ 9 + 10

= 0

Verify relations:

General form of a cubic equation: ax^3 + bx^2 + cx + d.

Now,

Consider Î± = 5, Î² = – 2 and Î³ = 1/3

Î± + Î² + Î³ = 5 â€“ 2 + 1/3 = 10/3 = -b/a

Î±Î² + Î²Î³ + Î±Î³ = 5 (-2) + (-2) (1/3) + (1/3) (5) = -9 = c/a

And, Î±Î²Î³ = 5 (-2) (1/3)= -10/3 = -d/a

Question 3: Find a cubic polynomial whose zeroes are 2, -3 and 4.

Solution:

General form of a cubic polynomial whose zeroes are a, b and c is:

x^3 â€“ (a + b + c) x^2 + (ab + bc + ca)x â€“ abc …(1)

To Find: Cubic polynomial whose zeroes are 2, -3 and 4

Let us say, a = 2, b = – 3 and c = 4

Putting the values of a, b and c in the equation (1) we get:

= x^3 â€“ (2 â€“ 3 + 4) x^2 + (-6 â€“ 12 + 8) x â€“ (- 24)

= x^3 â€“ 3x^2 â€“ 10x + 24

Which is required polynomial.

Question 4: Find a cubic polynomial whose zeroes are 1/2, 1 and -3.

Solution:

General form of a cubic polynomial whose zeroes are a, b and c is:

x^3 â€“ (a + b + c) x^2 + (ab + bc + ca)x â€“ abc …(1)

To Find: Cubic polynomial whose zeroes are 1/2, 1 and -3

Let us say, a = 1/2, b = 1 and c = -3

Putting the values of a, b and c in the equation (1) we get:

= x^3 â€“ (1/2 + 1 – 3) x^2 + (1/2 â€“ 3 â€“ 3/2)x â€“ (- 3/2)

= x^3 â€“ (-3/2) x^2 â€“ 4x + 3/2

= 2x^3 + 3x^2 â€“ 8x + 3

Which is required polynomial.

Question 5: Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time and the product of its zeroes as 5, -2 and -24 respectively.

Solution:

As we know, General form of a cubic polynomial whose zeroes are a, b and c is:

x^3 â€“ (a + b + c) x^2 + (ab + bc + ca)x â€“ abc

Also written as :

x^3 â€“ (Sum of the zeros) x^2 + (Sum of the product of the zeros taking two at a time) x â€“ (Product of Zeros) …(1)

Given:

Sum of the zeros = 5

Sum of the product of its zeros taken two at a time -2

Product of its zeros = â€“24

Putting these values in the equation (1), we get:

x^3 â€“ 5x^2 â€“ 2x + 24

Which is required polynomial.

Question 6: Find the quotient and remainder when

f(x) = x^3-3x^2+5x-3 is divided by g(x)= x^2-2

Solution:

Divide f(x) = x^3-3x^2+5x-3 by g(x)= x^2-2

Quotient = x – 3

Remainder = 7x – 9

Question 7: Find the quotient and remainder when

f(x) = x^4 â€“ 3x^2 + 4x + 5 is divided by g(x)= x^2 + 1 â€“ x

Solution:

f(x) = x^4 â€“ 3x^2 + 4x + 5 divide by g(x)= x^2 + 1 â€“ x

Quotient = x^2 + x – 3

Remainder = 8

Question 8: Find the quotient and remainder when

f(x) = x^4 – 5x + 6 is divided by g(x) = 2 â€“ x^2

Solution:

Quotient = -x^2 – 2

Remainder = -5x + 10

Question 9: By actual division, show that x2 â€“ 3 is a factor of 2x^4 + 3x^3 – 2x^2 â€“ 9x – 12.

Solution:

x2 â€“ 3 is a factor of 2x^4 + 3x^3 – 2x^2 â€“ 9x â€“ 12 only if remainder is zero.

Divide 2x^4 + 3x^3 – 2x^2 â€“ 9x â€“ 12 by x2 â€“ 3

Question 10: On dividing 3x^3 + x^2 + 2x + 5 is divided by a polynomial g(x), the quotient and remainder are (3x – 5) and (9x + 10) respectively. Find g(x).

Solution:

As per Division Rule:

Dividend = Quotient Ã— Divisor + Remainder …(i)

Given :

Dividend = 3x^3 + x^2 + 2x + 5

Quotient = 3x â€“ 5

Remainder = 9x + 10

(i) implies:

(3x^3 + x^2 + 2x + 5) / (3x â€“ 5) = g (x)

Therefore g(x) = x^2 + 2x + 1.

Question 11: Verify division algorithm for the polynomial f(x) = 8 + 20x + x^2 â€“ 6x^3 by g(x) = 2 + 5x – 3x^2.

Solution:

f(x) = -6x^3 + x^2 + 20x + 8

g(x) = -3x^2 + 5x + 2

Quotient = 2x + 3

Remainder = x + 2

Verification:

As per division rule:

Dividend = Quotient x Divisor + Remainder

After Putting the values, we get:

– 6x^3 + x^2 + 20x + 8 = (- 3x^2 + 5x + 2) (2x + 3) + (x + 2)

-6x^3 + x^2 + 20x + 8 = -6x^3 + x^2 + 20x + 8

Question 12: It is given that -1 is one of the zeroes of the polynomial x^3 + 2x^2 â€“ 11x â€“ 12 . Find all the zeroes of the given polynomial.

Solution:

Let f (x) = x^3 + 2x^2 â€’ 11x â€’ 12

Given: -1 is a zero of the polynomial, which means (x + 1) is a factor of f (x).

Dividing f (x) by (x + 1), we get

f (x) becomes:

f(x) = (x + 1) (x^2 + x â€“ 12)

= (x + 1) (x^2 + 4x â€“ 3x â€“ 12)

= (x + 1) (x – 3) (x + 4)

If f (x) = 0

• (x + 1) (x – 3) (x + 4) = 0

Either (x + 1) = 0 or (x â€“ 3) = 0 or (x + 4) = 0

x = -1 or x = 3 or x = – 4

Zeros of the polynomial are -1, 3 and -4

Question 13: If 1 and â€“2 are two zeroes of the polynomial (x^3 â€“ 4x^2 â€“ 7x + 10) , find its third zero.

Solution:

Let f(x) = x^3 â€“ 4x^2 â€“ 7x + 10

Given: 1 and â€“ 2 are the zeros of the given polynomial therefore (x â€“ 1) and (x + 2) is are factors of f (x).

Consequently, (x â€“ 1) (x + 2) = (x^2 + x â€“ 2) is a factor of f(x).

Divide x^3 â€“ 4x^2 â€“ 7x + 10 by (x^2 + x â€“ 2):

Put f (x) = 0

(x^2 + x â€“ 2) (x â€“ 5) = 0

(x â€“ 1) (x + 2) (x â€“ 5) = 0

x = 1 or x = – 2 or x = 5

Hence, the third zero is 5.

Question 14: If 3 and â€“3 are two zeroes of the polynomial (x^4 + x^3 â€“ 11x^2 â€“ 9x + 18), find all the zeroes of the given polynomial.

Solution:

Let f(x) = x^4 + x^3 â€“ 11x^2 â€“ 9x + 18

Given: 3 and â€“ 3 are the zeros of the polynomial, (x + 3) and (x – 3) are factors of f (x), and

Consequently, (x â€“ 3) (x + 3) = (x^2 â€“ 9) is a factor of f (x)

Divide f(x) by (x^2 â€“ 9) we get:

Put f(x) = 0

(x^2 + x â€“ 2) (x^2 â€“ 9) = 0

(x â€“ 1) (x + 2) (x â€“ 3) (x + 3) = 0

x = 1 or x = – 2 or x = 3 or x = – 3

Hence, all the zeros of the given polynomial are 1, -2, 3 and -3

Question 15: If 2 and -2 are two zeroes of the polynomial (x^4 + x^3 â€“ 34x^2 â€“ 4x + 120), find all the zeroes of the given polynomial.

Solution:

Let us assume f (x) = x^4 + x^3 â€“ 34x^2 â€“ 4x + 120

Given: 2 and â€“ 2 are the zeros of the polynomial

(x – 2) and (x + 2) are factors of f(x) and

(x â€“ 2) (x + 2) = (x^2 â€“ 4) is a factor of f (x), and

Divide f(x) by (x^2 â€“ 4) we get:

Set f (x) = 0

(x^2 + x â€“ 30) (x^2 â€“ 4) = 0

(x^2 + 6x â€“ 5x â€“ 30) (x â€“ 2)(x + 2) = 0

[x (x + 6) â€“ 5 (x + 6)] (x â€“ 2) (x + 2) = 0

(x â€“ 5) (x + 6) (x â€“ 2) (x + 2) = 0

either x = 5 or x = – 6 or x = 2 or x = – 2

Hence, all the zeros of the given polynomial are 2, -2, 5 and -6.

Question 16: Find all the zeroes of x^4 + x^3 – 23x^2 – 3x + 60, if it is given that two of its zeroes are âˆš3 and â€“ âˆš3 .

Solution:

Let f(x) = x^4 + x^3 – 23x^2 – 3x + 60

Given: âˆš3 and â€“âˆš3 are the zeros of the polynomial.

(x – âˆš3) and (x – âˆš3) are factors of f(x)

So, (x â€“ âˆš3) (x + âˆš3) = (x^2 â€“ 3) is a factor of f(x)

Divide f(x) by (x^2 â€“ 3):

Set f(x) = 0

(x^2 + x â€“ 20) (x^2 â€“ 3) = 0

(x^2 + 5x â€“ 4x â€“ 20) (x^2 â€“ 3) =0

[x (x + 5) â€“ 4 (x + 5)] (x2 â€“ 3)=0

(x â€“ 4) (x + 5) (x â€“ ) (x + ) = 0

either x = 4 or x = – 5 or x = âˆš3 or x = – âˆš3

Hence, all the zeros of the given polynomial are âˆš3, -âˆš3, 4 and – 5

Question 17: Find all the zeroes of (2x^4 – 3x^3 – 5x^2 + 9x – 3) , it is being given that two of its zeroes are âˆš3 and – âˆš3.

Solution:

Let f(x) = 2x^4 – 3x^3 – 5x^2 + 9x – 3

Given: âˆš3 and â€“âˆš3 are the zeros of the polynomial.

(x – âˆš3) and (x – âˆš3) are factors of f(x)

So, (x â€“ âˆš3) (x + âˆš3) = (x^2 â€“ 3) is a factor of f(x)

Divide f(x) by (x^2 â€“ 3):

Set f(x) = 0

(2x^2 – 3x + 1) (x^2 â€“ 3) = 0

2x^2 – 3x^2 â€“ 5x^2 + 9x – 3 = 0

(x^2 â€“ 3) (2x^2 â€“ 3x + 1) = 0

(x^2 – 3) (2x^2 â€“ 2x â€“ x + 1) (2x â€“ 1) (x – 1) = 0

(x â€“ âˆš3) (x + âˆš3) (2x â€“ 1) (x – 1) = 0

either x = âˆš3 or x = -âˆš3 or x = 1/2 or x = 1

Hence, all the zeros of the given polynomial are âˆš3 , -âˆš3 , 1/2 and 1

Question 18: Obtain all other zeroes of (x^4 + 4x^3 â€“ 2x^2 â€“ 20x â€“ 15) if two of its zeroes are âˆš5 and â€“âˆš5.

Solution:

Let f(x) = x^4 + 4x^3 â€“ 2x^2 â€“ 20x â€“ 15

(x-âˆš5) and (x-âˆš5) are the factors of f(x)

and (x-âˆš5) (x + âˆš5) = (x^2 â€“ 5) is a factor of f (x)

Divide f(x) by (x^2 â€“ 5) we get:

Set f(x) = 0

x^4 + 4x^3 â€“ 2x^2 â€“ 20x â€“ 15 = 0

(x^2 â€“ 5) (x^2 + 4x + 3) = 0

(x -âˆš5) (x + âˆš5) (x + 1) (x + 3) = 0

either x = âˆš5 or x = – âˆš5 or x = – 1 or x = – 3

Hence, all the zeros of the given polynomial are âˆš5, -âˆš5, – 1 and – 3.

Question 19: Find all the zeroes of polynomial (2x^4 â€“ 11x^3 + 7x^2 + 13x â€“ 7) , it being given that two of its zeroes are (3 + âˆš2) and (3 â€“ âˆš2).

Solution:

Let f(x) = 2x^4 â€“ 11x^3 + 7x^2 + 13x â€“ 7

Given: (3 + âˆš2) and (3- âˆš2) are the zeros of f(x).

So (x – (3 + âˆš2)) and (x – (3 – âˆš2)) are factors of f(x)

and (x – (3 + âˆš2))(x – (3 – âˆš2)) = x^2 – 6x + 7 is a factor of f(x)

Divide f(x) by x^2 – 6x + 7 we get:

Set f(x) = 0

2x^4 â€“ 11x^3 + 7x^2 + 13x â€“ 7 = 0

(x^2 – 6x + 7) (2x^2 + x â€“ 1) = 0

(x + 3 + âˆš2) (x + 3 -âˆš2) (2x â€“ 1) (x + 1) = 0

x = – 3 – âˆš2 or x = – 3 + âˆš2 or x = 1/2 or x = – 1

Hence, all the zeros of the given polynomial are (-3 -âˆš2), (-3 + âˆš2), 1/2 and â€“ 1.

## RS Aggarwal Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2B

Class 10 Maths Chapter 2 Polynomials Exercise 2B is based on polynomials and relation between the zeros and coefficients of a quadratic Polynomial.