RS Aggarwal Solutions for Class 10 Maths Chapter 2 Polynomials

Get RS Aggarwal Solutions for Class 10 Chapter 2 Polynomials here. The subject experts at BYJU’S have solved each question of RS Aggarwal meticulously to help the students in solving any question from the book. Practice Polynomials RS Aggarwal questions and become master on the concepts. All solutions are explained using step-by-step approach. Download RS Aggarwal Maths Class 10 Solutions and score more in your exams.

 

Access Other Exercise Solutions of RS Aggarwal Class 10 Maths Chapter 2:

Get detailed solutions for all the questions listed under the below exercises:

Exercise 2A Solutions: 21 Questions (Short Answers)

Exercise 2B Solutions: 19 Questions (Short Answers)

Exercise 2C Solutions: 25 Questions (Short Answers)

Exercise 2A Page No: 48

 

Solution:

1. x^2 + 7x + 12

Let f(x) = x^2 + 7x + 12

f(x) = x^2 + 4x + 3x + 12

f(x) = x(x+4) + 3(x+4)

f(x) = (x+4)(x+3)

To find the zeroes, set f(x) = 0, then

either (x + 4) = 0 or (x + 3) = 0

x = −4 or x = −3

Again,

Sum of zeroes = (-4 – 3) = -7 = -7/1

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = 12 = 12/1

= c/a

= Constant term / Coefficient of x^2

2. x^2 – 2x – 8

Let f(x) = x^2 ˗ 2x ˗ 8

= x^2 ˗ 4x + 2x ˗ 8

= x(x ˗ 4) + 2(x ˗ 4)

= (x ˗ 4) (x + 2)

To find the zeroes, set f(x) = 0, then

either (x ˗ 4) = 0 or (x+2) = 0

x = 4 or x = -2

Again,

Sum of zeroes = (4 – 2) = 2 = 2/1

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = (4) (-2) = -8/1

= c/a

= Constant term / Coefficient of x^2

3. x^2 + 3x ˗ 10

Let f(x) = x^2 + 3x ˗ 10

= x^2 + 5x ˗ 2x ˗ 10

= x(x + 5) ˗ 2(x + 5)

= (x ˗ 2) (x + 5)

To find the zeroes, set f(x) = 0, then

either x ˗ 2 = 0 or x + 5 = 0

⇒ x = 2 or x = −5.

So, the zeroes of f(x) are 2 and −5.

Again,

Sum of zeroes = 2 + (-5) = -3 = (-3)/1

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = (2)(-5) = -10 = (-10)/1

= c/a

= Constant term / Coefficient of x^2

4. 4x^2 ˗ 4x – 3

Let f(x) = 4x^2 ˗ 4x ˗ 3

= 4x^2 ˗ (6x ˗ 2x) ˗ 3

= 4x^2 ˗ 6x + 2x ˗ 3

= 2x (2x ˗ 3) + 1(2x ˗ 3)

= (2x + 1) (2x ˗ 3)

To find the zeroes, set f(x) = 0

(2x + 1) (2x ˗ 3)= 0

2x + 1 = 0 or 2x ˗ 3 = 0

x = -1/2 or x = 3/2

Again,

Sum of zeroes = (-1/2) + (3/2) = (-1+3)/2 = 2/2

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = (-1/2)(3/2) = (-3)/4

= c/a

= Constant term / Coefficient of x^2

5. 5x^2 ˗ 4 ˗ 8x

Let f(x) = 5x^2 ˗ 4 ˗ 8x

= 5x^2 ˗ 8x ˗ 4

= 5x^2 ˗ (10x ˗ 2x) ˗ 4

= 5x^2 ˗ 10x + 2x ˗ 4

= 5x (x ˗ 2) + 2(x ˗ 2)

= (5x + 2) (x ˗ 2)

To find the zeroes, set f(x) = 0

(5x + 2) (x ˗ 2) = 0

5x + 2 = 0 or x ˗ 2 = 0

x = (−2)/5 or x = 2

Again,

Sum of zeroes = (-2)/5 + 2 = (-2+10)/5 = 8/5

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = (-2/5) x 2 = (-4)/5

= c/a

= Constant term / Coefficient of x^2

6. 2 √3 x^2 – 5x + √3

Let f(x) = 2 √3 x^2 – 5x + √3

= 2 √3 x^2 – 2x – 3x + √3

= 2x(√3x-1) – √3(√3x-1)

To find the zeroes, set f(x) = 0

(√3x – 1) or (2x – √3) = 0

x = 1/√3 = √3/3 or x = √3/2

x = √3/3 or x = √3/2

Again,

Sum of zeroes = 1/√3 + √3/2 = 5/2√3

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = 1/√3 x √3/2 = √3/2√3

= c/a

= Constant term / Coefficient of x^2

7. 2x^2 – 11x + 15

Let f(x) = 2x^2 ˗ 11x + 15

= 2x^2 ˗ (6x + 5x) + 15

= 2x^2 ˗ 6x ˗ 5x + 15

= 2x (x ˗ 3) ˗ 5 (x ˗ 3)

= (2x ˗ 5) (x ˗ 3)

To find the zeroes, set f(x) = 0

(2x ˗ 5) (x ˗ 3) = 0

2x ˗ 5= 0 or x ˗ 3 = 0

x = 5/2 or x = 3

Again,

Sum of zeroes = 5/2 + 3 = (5+6)/2 = 11/2

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = 5/2 x 3 = 15/2

= c/a

= Constant term / Coefficient of x^2

8. 4x^2 ˗ 4x + 1

Let f(x) = 4x^2 ˗ 4x + 1

= (2x^2) – 2(2x)(1) + (1)^2

= (2x – 1)^2

To find the zeroes, set f(x) = 0

(2x – 1)^2 = 0

x = 1/2 or x = 1/2

Again,

Sum of zeroes = 1/2+1/2=1=1/1

= -b/a

= (-Coefficient of x) / (Cofficient of x^2)

Product of zeroes = 1/2 x 1/2=1/4

= c/a

= Constant term / Coefficient of x^2

9. (x^2 – 5)

Let f(x) = (x^2 – 5)

= (x^2-(√5)^2)

= (x + √5)(x – √5)

To find the zeroes, set f(x) = 0

(x + √5)(x – √5)=0

x = -√5 or x = √5

Again,

Sum of zeroes = -√5 + √5 = 0/1

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = -√5 + √5 = -5 = -5/1

= c/a

= Constant term / Coefficient of x^2

10. (8x^2 ˗ 4)

Let f(x) = 8x^2 – 4

= 4 ((√2x)^2 – (1)^2)

= 4(√2x + 1)(√2x – 1)

To find the zeroes, set f(x) = 0

(√2x + 1)(√2x – 1) = 0

(√2x + 1) = 0 or (√2x – 1) = 0

x = (-1)/√2 or x = 1/√2

So, the zeroes of f(x) are (-1)/√2 and x = 1/√2

Again,

Sum of zeroes = -1/√2 + 1/√2 = (-1+1)/√2 = 0

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = -1/√2 x 1/√2 = -1/2 = -4/8

= c/a

= Constant term / Coefficient of x^2

11. (5y^2 + 10y)

Let f(x) = (5y^2 + 10y)

= 5y (y + 2)

To find the zeroes, set f(x) = 0

5y (y + 2) = 0

y = 0 or y = -2

So, the zeroes of f(x) are 0 and -2

Again,

Sum of zeroes = -2 + 0 = -2 = -10/5

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = -2 x 0 = 0

= c/a

= Constant term / Coefficient of x^2

12. (3x^2 ˗ x ˗ 4)

Let f(x) = 3x^2 ˗ x ˗ 4

3x^2 ˗ 4x + 3x ˗ 4

= x(3x ˗ 4) + 1 (3x ˗ 4)

= (3x ˗ 4) (x + 1)

To find the zeroes, set f(x) = 0

(3x ˗ 4) = 0 or (x + 1) = 0

⇨ x = 4/3 or x=-1

So, the zeroes of f(x) are 4/3 and x=-1

Again,

Sum of zeroes = 4/3 + (-1) = 1/3

= -b/a

= (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = 4/3 + (-1) = -4/3

= c/a

= Constant term / Coefficient of x^2

13. 

 

Solution:

Let ∝ = 2 and β = -6

Sum of the zeroes = (∝ + β) = 2 – 6 = -4

Product of the zeroes, 𝛼𝛽 = 2(-6) = -12

Required quadratic polynomial is

⇨ x^2 – (∝+β)x + ∝β = x^2 – (-4)x – 12

= x^2 + 4x – 12

And,

Sum of the zeroes = – 4 = -4/1 = (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = -12 = -12/1 = Constant term / Coefficient of x^2

14.

 

Solution:

Let ∝ = 2/3 and β = -1/4

Sum of the zeroes = (∝ + β) = 2/3 + -1/4 = 5/12

Product of the zeroes, 𝛼𝛽 = 2/3 x -1/4 = -1/6

Required quadratic polynomial is

⇨ x^2 – (∝+β)x + ∝β = x^2 – (5/12)x – (-1/6)

= 1/12(12x^2 – 5x – 2)

And,

Sum of the zeroes = 5/12 = (-Coefficient of x)/(Cofficient of x^2)

Product of zeroes = -1/6 = Constant term / Coefficient of x^2

15. 

 

Solution:

Given : Sum of zeroes = (∝ + β) = 8

Product of the zeroes = 𝛼𝛽 = 12

Required quadratic polynomial is

x^2 – (∝+β)x + ∝β = x^2 – (8)x + 12

Now, find the zeroes of the above polynomial.

Let f(x) = x^2 – (8)x + 12

= x^2 – 6x – 2x + 12

= (x -6)(x – 2)

Substitute f(x) = 0.

either (x -6) = 0 or (x – 2) = 0

⇨ x = 6 or x = 2

2 and 6 are the zeroes of the polynomial.

16. 

 

Solution:

Given : Sum of zeroes = (∝ + β) = 0

Product of the zeroes = 𝛼𝛽 = -1

Required quadratic polynomial is

x^2 – (∝+β)x + ∝β = x^2 – (0)x – 1

= x^2 – 1

Now, find the zeroes of the above polynomial.

Let f(x) = x^2 – 1

= x^2 – 1^2

= (x – 1)(x + 1)

Substitute f(x) = 0.

either (x – 1) = 0 or (x + 1) = o

⇨ x = 1 or x = -1

1 and -1 are the zeroes of the polynomial.

17. 

 

Solution:

Given : Sum of zeroes = (∝ + β) = 5/2

Product of the zeroes = 𝛼𝛽 = 1

Required quadratic polynomial is

x^2 – (∝+β)x + ∝β = x^2 – (5/2)x + 1

= 1/2(2x^2 – 5x + 2)

Now, find the zeroes of the above polynomial.

Let f(x) = 1/2(2x^2 – 5x + 2)

= 1/2(2x^2 – 4x – x + 2)

= 1/2(2x(x – 2) – (x – 2))

= 1/2( (2x – 1)(x – 2) )

Substitute f(x) = 0.

1/2( (2x – 1)(x – 2) ) = 0

either (2x – 1) = 0 or (x – 2) = o

⇨ x = 1/2 or x = 2

1/2 and 2 are the zeroes of the polynomial.

18. 

 

Solution:

Given : Sum of zeroes = (∝ + β) = √2

Product of the zeroes = 𝛼𝛽 = 1/3

Required quadratic polynomial is

x^2 – (∝+β)x + ∝β = x^2 – (√2)x + 1/3

= x^2 – √2x + 1/3

19. 

 

Solution:

Given roots are: x = 2/3 and x = -3

and quadratic equation ax^2 + 7x + b = 0

Since x = 2/3 and x = -3 are roots of the above quadratic equation

Hence, will satisfy the given equation.

Step 1: At x = 2/3

a(2/3)^2 + 7(2/3) + b=0

4/9 a + 14/3 + b = 0

4a + 42 + 9b = 0 ………….equation (1)

Step 2: At x = –3

a(-3)^2+7(-3)+b=0

9a – 21 + b = 0 ………….equation (2)

Step 3: Solving equation (1) and equation (2), we get

a = 3, b = –6

20.

Solution:

Given: (x + a) is a factor of polynomial 2x^2 + 2ax + 5x + 10.

So, we have

x + a = 0

or x = –a , will satisfy the given polynomial.

Therefore, we will have

2 (–a)^2 + 2a(–a) + 5(–a) + 10 = 0

2a^2 –2a^2 – 5a + 10 = 0

– 5a = – 10

a = 2

The value of a is 2.

21.

Solution:

Given: x = 2/3 is one of the zero of polynomial 3x^3 + 16x^2 + 15x – 18

Now, we have

x = 2/3

or x – 2/3 = 0

To find other zeroes, let us divide the polynomial 3x^3 + 16x^2 + 15x – 18 by x – 2/3.

rs aggarwal chapter 2 ex a img 1

So, the quotient is 3x^2 + 18x + 27

Again, set 3x^2 + 18x + 27 = 0 to find the other zeroes.

3x^2 + 18x + 27 = 0

3x^2 + 9x + 9x + 27 = 0

3x(x + 3) + 9(x + 3) = 0

(x + 3) (3x + 9) = 0

Either (x + 3) = 0 or (3x + 9) = 0

x = -3 or x = -3


Exercise 2B Page No: 58

Question 1: 

 

Solution:

Let f(x) = x^3 – 2x^2 – 5x + 6

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

Therefore,

f(3) = (3)^3 – 2(3)^2 – 5(3) + 6

= 27 – 18 – 15 + 6

= 0

f(-2) = (-2)^3 – 2(-2)^2 – 5(-2) + 6

= -8 – 8 + 10 + 6

= 0

f(1) = (1)^3 – 2 (1)^2 – 5 (1) + 6

= 1 – 2 – 5 + 6

= 0

Verify relations:

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

Now,

Consider α = 3, β = – 2 and γ = 1

α + β + γ = 3 – 2 + 1 = 2 = -b/a

αβ + βγ + αγ = 3 (-2) + (-2) (1) + 1 (3) = -5 = c/a

And, αβγ = 3 (-2)(1) = -6 = -d/a

Question 2: 

 

Solution:

Let f(x) = 3x^3 – 10x^2 – 27x + 10

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

Therefore,

f(5) = 3(5)^3 – 10(5)^2 – 27(5) + 10

= 3 × 125 – 250 – 135 + 10

= 0

f(-2) = 3(-2)^3 – 10(-2)^2 – 27(-2) + 10

= – 24 – 40 + 54 + 10

= 0

f(1/3) = 3 (1/3)^3 – 10(1/3)^2 – 27(1/3) + 10

= 1/9 – 10/9 – 9 + 10

= 0

Verify relations:

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

Now,

Consider α = 5, β = – 2 and γ = 1/3

α + β + γ = 5 – 2 + 1/3 = 10/3 = -b/a

αβ + βγ + αγ = 5 (-2) + (-2) (1/3) + (1/3) (5) = -9 = c/a

And, αβγ = 5 (-2) (1/3)= -10/3 = -d/a

Question 3: 

 

Solution:

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

x^3 – (a + b + c) x^2 + (ab + bc + ca)x – 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 – (2 – 3 + 4) x^2 + (-6 – 12 + 8) x – (- 24)

= x^3 – 3x^2 – 10x + 24

Which is required polynomial.

Question 4:

Solution:

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

x^3 – (a + b + c) x^2 + (ab + bc + ca)x – 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 – (1/2 + 1 – 3) x^2 + (1/2 – 3 – 3/2)x – (- 3/2)

= x^3 – (-3/2) x^2 – 4x + 3/2

= 2x^3 + 3x^2 – 8x + 3

Which is required polynomial.

Question 5:

Solution:

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

x^3 – (a + b + c) x^2 + (ab + bc + ca)x – abc

Also written as :

x^3 – (Sum of the zeros) x^2 + (Sum of the product of the zeros taking two at a time) x – (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 = –24

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

x^3 – 5x^2 – 2x + 24

Which is required polynomial.

Question 6:

 

Solution:

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

rs aggarwal chapter 2 ex b img 1

Quotient = x – 3

Remainder = 7x – 9

Question 7:

 

Solution:

f(x) = x^4 – 3x^2 + 4x + 5 divide by g(x)= x^2 + 1 – x

rs aggarwal chapter 2 ex b img 2

Quotient = x^2 + x – 3

Remainder = 8

Question 8:

 

Solution:

rs aggarwal chapter 2 ex b img 3

Quotient = -x^2 – 2

Remainder = -5x + 10

Question 9: 

 

Solution:

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

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

rs aggarwal chapter 2 ex b img 4

Question 10: 

 

Solution:

As per Division Rule:

Dividend = Quotient × Divisor + Remainder …(i)

Given :

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

Quotient = 3x – 5

Remainder = 9x + 10

(i) implies:

(3x^3 + x^2 + 2x + 5) / (3x – 5) = g (x)

rs aggarwal chapter 2 ex b img 5

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

Question 11: 

 

Solution:

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

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

rs aggarwal chapter 2 ex b img 6

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: 

 

Solution:

Let f (x) = x^3 + 2x^2 ‒ 11x ‒ 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

rs aggarwal chapter 2 ex b img 7

f (x) becomes:

f(x) = (x + 1) (x^2 + x – 12)

= (x + 1) (x^2 + 4x – 3x – 12)

= (x + 1) (x – 3) (x + 4)

If f (x) = 0

  • (x + 1) (x – 3) (x + 4) = 0

Either (x + 1) = 0 or (x – 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:

Solution:

Let f(x) = x^3 – 4x^2 – 7x + 10

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

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

Divide x^3 – 4x^2 – 7x + 10 by (x^2 + x – 2):

rs aggarwal chapter 2 ex b img 8

Put f (x) = 0

(x^2 + x – 2) (x – 5) = 0

(x – 1) (x + 2) (x – 5) = 0

x = 1 or x = – 2 or x = 5

Hence, the third zero is 5.

Question 14: 

 

Solution:

Let f(x) = x^4 + x^3 – 11x^2 – 9x + 18

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

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

Divide f(x) by (x^2 – 9) we get:

rs aggarwal chapter 2 ex b img 9

Put f(x) = 0

(x^2 + x – 2) (x^2 – 9) = 0

(x – 1) (x + 2) (x – 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: 

 

Solution:

Let us assume f (x) = x^4 + x^3 – 34x^2 – 4x + 120

Given: 2 and – 2 are the zeros of the polynomial

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

(x – 2) (x + 2) = (x^2 – 4) is a factor of f (x), and

Divide f(x) by (x^2 – 4) we get:

rs aggarwal chapter 2 ex b img 10

Set f (x) = 0

(x^2 + x – 30) (x^2 – 4) = 0

(x^2 + 6x – 5x – 30) (x – 2)(x + 2) = 0

[x (x + 6) – 5 (x + 6)] (x – 2) (x + 2) = 0

(x – 5) (x + 6) (x – 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: 

 

Solution:

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

Given: √3 and –√3 are the zeros of the polynomial.

(x – √3) and (x – √3) are factors of f(x)

So, (x – √3) (x + √3) = (x^2 – 3) is a factor of f(x)

Divide f(x) by (x^2 – 3):

rs aggarwal chapter 2 ex b img 11

Set f(x) = 0

(x^2 + x – 20) (x^2 – 3) = 0

(x^2 + 5x – 4x – 20) (x^2 – 3) =0

[x (x + 5) – 4 (x + 5)] (x2 – 3)=0

(x – 4) (x + 5) (x – ) (x + ) = 0

either x = 4 or x = – 5 or x = √3 or x = – √3

Hence, all the zeros of the given polynomial are √3, -√3, 4 and – 5

Question 17: 

 

Solution:

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

Given: √3 and –√3 are the zeros of the polynomial.

(x – √3) and (x – √3) are factors of f(x)

So, (x – √3) (x + √3) = (x^2 – 3) is a factor of f(x)

Divide f(x) by (x^2 – 3):

rs aggarwal chapter 2 ex b img 12

Set f(x) = 0

(2x^2 – 3x + 1) (x^2 – 3) = 0

2x^2 – 3x^2 – 5x^2 + 9x – 3 = 0

(x^2 – 3) (2x^2 – 3x + 1) = 0

(x^2 – 3) (2x^2 – 2x – x + 1) (2x – 1) (x – 1) = 0

(x – √3) (x + √3) (2x – 1) (x – 1) = 0

either x = √3 or x = -√3 or x = 1/2 or x = 1

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

Question 18:

Solution:

Let f(x) = x^4 + 4x^3 – 2x^2 – 20x – 15

(x-√5) and (x-√5) are the factors of f(x)

and (x-√5) (x + √5) = (x^2 – 5) is a factor of f (x)

Divide f(x) by (x^2 – 5) we get:

rs aggarwal chapter 2 ex b img 13

Set f(x) = 0

x^4 + 4x^3 – 2x^2 – 20x – 15 = 0

(x^2 – 5) (x^2 + 4x + 3) = 0

(x -√5) (x + √5) (x + 1) (x + 3) = 0

either x = √5 or x = – √5 or x = – 1 or x = – 3

Hence, all the zeros of the given polynomial are √5, -√5, – 1 and – 3.

Question 19: 

 

Solution:

Let f(x) = 2x^4 – 11x^3 + 7x^2 + 13x – 7

Given: (3 + √2) and (3- √2) are the zeros of f(x).

So (x – (3 + √2)) and (x – (3 – √2)) are factors of f(x)

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

Divide f(x) by x^2 – 6x + 7 we get:

rs aggarwal chapter 2 ex b img 14

Set f(x) = 0

2x^4 – 11x^3 + 7x^2 + 13x – 7 = 0

(x^2 – 6x + 7) (2x^2 + x – 1) = 0

(x + 3 + √2) (x + 3 -√2) (2x – 1) (x + 1) = 0

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

Hence, all the zeros of the given polynomial are (-3 -√2), (-3 + √2), 1/2 and – 1.


Exercise 2C Page No: 59

QUESTION 1:

 

SOLUTION:

Given: (2 + √3) is one of zero of polynomial x^2 ‒ 4x + 1.

To find: Other zero

Since given polynomial is a quadratic so it has only two zeros.

Let other zero be x.

Now,

Sum of zeros = -(coefficient of x)/(coefficient of x^2)

x + (2 + √3) = -(-4/1) = 4

x + 2 + √3 = 4

or x = 2 – √3

Hence, the other zero is (2 -√3).

QUESTION 2: 

 

SOLUTION:

Let f(x) = x^2 + x ‒ p (p + 1)

Above polynomial can be written as,

= x^2 + (p + 1) x – px – p (p + 1)

= x (x + (p + 1)) – p (x + (p + 1))

= (x – p)(x + (p + 1))

To find the zeroes of f(x), put f(x) = 0

(x – p)(x + (p + 1)) = 0

either (x – p) = 0 or (x + (p + 1)) = 0

x = p or x = – (p + 1)

Hence, the zeros of the given polynomial are p and – (p + 1)

QUESTION 3

 

 

SOLUTION:

Let f(x) = x^2 – 3x – m (m + 3)

Above polynomial can be written as,

f(x) = x^2 – (m + 3)x + mx – m(m + 3)

= x(x – m – 3) + m(x – m – 3)

= (x – m – 3)(x + m)

To find the zeroes of f(x), put f(x) = 0

(x – m – 3)(x + m) = 0

Either x – m – 3 = 0 or x + m = 0

x = m + 3 or x = -m

Required Zeros are (m + 3), -m

QUESTION 4

 

 

SOLUTION:

Given : α + β= 6 = sum of zeros and

αβ = 4 = Product of zeroes

We know that, if α and β are the zeros of the polynomial then the quadratic polynomial can be x^2 – (α + β) x + αβ

Now substituting the values, we get

x^2 – 6x + 4, which is required polynomial.

QUESTION 5

 

 

SOLUTION:

Given: 2 is one of zeroes of the polynomial kx^2 + 3x + k , which means x = 2 will satisfy it.

k(2)^2 + 3( 2) + k = 0

4k + 6 + k= 0

5k + 6 = 0

k = -6 / 5

The value of k is -6 / 5

QUESTION 6

 

 

SOLUTION:

Given: 3 is one of zeroes of the polynomial 2x^2 + x + k , which means x = 3 will satisfy it.

2x^2 + x + k = 0

2(3)^2 + 3 + k = 0

18 + 3+ k = 0

k = -21

the value of k is -21

QUESTION 7

 

SOLUTION:

Given: -4 is one of zeroes of the polynomial x^2 ‒ x- (2k + 2), which means x = -4 will satisfy it.

x^2 – x – (2k + 2) = 0

(-4)^2 – (-4) – 2k – 2 = 0

16 + 4 – 2k – 2 = 0

-2k + 18 = 0

k = 9

The value of k is 9.

QUESTION 8

 

 

SOLUTION:

Given: 1 is one of zeroes of the polynomial ax^2 ‒ 3 (a ‒ 1) x ‒ 1 , which means x = 1 will satisfy it.

a(1)^2 – 3(a – 1) x 1 – 1 = 0

a – 3a + 3 – 1 = 0

-2a + 2 = 0

a = 1

The value of a is 1.

QUESTION 9

 

SOLUTION:

Given: -2 is one of zeroes of the polynomial 3x^2 + 4x + 2k , which means x = -2 will satisfy it.

3(-2)^2 + 4(-2) + 2k = 0

12 – 8 + 2k = 0

4 + 2k=0

k = -2

The value of k is -2.

QUESTION 10

 

 

SOLUTION:

Let f(x) = x^2 – x – 6

= x^2 – 3x + 2x – 6

= x(x – 3) + 2(x – 3)

= (x – 3)(x + 2)\

To find the zeros of f(x), let sat f(x) = 0, we get

(x – 3)(x + 2) = 0

Either x – 3 = 0 or x + 2 = 0

x = 3 or x = -2

Therefore, 3, -2 are zeros.

QUESTION 11

 

SOLUTION:

Given: Sum of zeros of polynomial is kx^2 – 3x + 5 is 1.

We know that,

Sum of zeros = -(coefficient of x)/(coefficient of x^2) = -(-3) / k = 3/k

From above results, we get

3/k = 1

Or k = 3

QUESTION 12

 

 

SOLUTION:

Given: Product of zeros of polynomial is x^2 ‒ 4x + k is 3.

We know that,

Product of zeros = (constant term)/(coefficient of x^2) = k/1 = k

From above results, we get

K = 3

QUESTION 13

 

 

SOLUTION:

Given: (x + a) is a factor of 2x^2 + (2a + 5) x + 10.

Which shows that one of the zeros of the given polynomial is x + a = 0, i.e. –a.

Now, x = -a satisfy the given polynomial.

2 (-a)^2 + (2a + 5)(-a) + 10 = 0

2a^2 – 2a^2 – 5a + 10 = 0

5a = 10

a = 2

QUESTION 14

 

 

SOLUTION:

(a – b), a, (a + b) are the zeros of 2x^3 – 6x^2 + 5x – 7

Sum of zeros = -(coefficient of x)/(coefficient of x^2)

a – b + a + a + b = -(-6) / 2 = 3

3a = 3

or a = 1

QUESTION 15

 

 

SOLUTION:

On dividing x^3 + x^2 ‒ ax + b by (x^2 ‒ x):

rs aggarwal chapter 2 ex c img 1

Remainder = (2 – a) x + b

Since given polynomial is divisible by x^2 – x (given)

So remainder should be zero.

Therefore,

(2 – a) x + b = 0

pnly possible if, 2 – a = 0 and b = 0

a = 2 and b = 0

Hence, the values are : a = 2, b = 0

QUESTION 16

 

SOLUTION:

Given: α and β are the zeros of polynomial 2x^2 + 7x + 5

α + β = Sum of zeros = -(coefficient of x)/(coefficient of x^2) = -7/2

αβ = Product of zeros = (constant term)/(coefficient of x^2) = 5/2

α + β + αβ = (α + β) + αβ = -7/2 + 5/2 = -1

QUESTION 17

 

 

SOLUTION:

Division algorithm for polynomials states that: f(x) = q(x) g(x) + r(x)

Where f(x) and g(x) are any two polynomials with g(x) ≠ 0.

After dividing f(x) by g(x), we get two another polynomials, q(x) and r(x)

where r (x) = 0 and degree of r(x) < degree of g(x)

QUESTION 18

 

 

SOLUTION:

Sum of zeros = -1/2

Product of zeros = -3

We know that, general form of a Polynomial:

x^2 – (Sum of zeros) x + product of zeros

This implies: x^2 – (-1/2) + (-3)

x^2 + 1/2x – 3

QUESTION 19

 

SOLUTION:

To find the zeros of the quadratic polynomial, put f(x) = 0

6x^2 – 3 = 0

3 (2x^2 – 1) = 0

2x^2 = 1

x^2 = 1/2

or x = ±1/√2

Zeros of the given polynomial are 1/√2 and -1/√2.

QUESTION 20

 

 

SOLUTION:

f(x) = 4√3x^2 + 5x ‒ 2√3

To find the zeros of the quadratic polynomial, put f(x) = 0

4√3x^2 + 5x ‒ 2√3 = 0

4√3 x^2 + 8x – 3x – 2√3 = 0

4x (√3 x + 2) – √3 (√3 x + 2) = 0

(4x – √3) (√3 x + 2) = 0

Either (√3 x + 2) = 0 or (4x – √3) = 0

x = -2/√3 or x = √3/4

QUESTION 21

 

 

SOLUTION:

Given: α and β are zeroes of f(x) = x^2 ‒ 5x + k

and α – β = 1

We know that,

α + β = Sum of zeros = -(coefficient of x)/(coefficient of x^2) = 5

α β = Product of zeros = (constant term)/(coefficient of x^2) = k

Now solving α – β = 1 and α + β = 5, we get:

α = 3 and β = 2

Again: α β = k (putting value of α and β)

k = 6.

QUESTION 22

 

SOLUTION:

Given: α and β are zeroes of f(x) = 6x^2 + x – 2

To find: (α/ β + β / α)

α + β = Sum of zeros = -(coefficient of x)/(coefficient of x^2) = -1/6

α β = Product of zeros = (constant term)/(coefficient of x^2) = -1/3

Now,

(α/ β + β / α) = {(α + β)^2 – 2αβ) / αβ }

= (1/36 + 2/3) / (-1/3)

= -25/12

QUESTION 23

 

 

SOLUTION:

Given: α and β are zeroes of f(x) = 5x^2 ‒ 7x + 1

To find: (1/α + 1/β)

α + β = Sum of zeros = -(coefficient of x)/(coefficient of x^2) = 7/5

α β = Product of zeros = (constant term)/(coefficient of x^2) = 1/5

Now,

(1/α + 1/β) = (α + β) / αβ

= 7/5 x 5/1

= 7

QUESTION 24

 

 

SOLUTION:

Given: α and β are zeroes of f(x) = x^2 + x – 2

To find: (1/α – 1/β)

α + β = Sum of zeros = -(coefficient of x)/(coefficient of x^2) = -1

α β = Product of zeros = (constant term)/(coefficient of x^2) = -2

Now,

(1/α – 1/β) = (β – α)^2) / αβ

= (β + α)^2 – 4αβ) / (αβ)^2

= 9/ 4

QUESTION 25

 

 

SOLUTION:

Given: Zeros of the polynomial x^3 – 3x^2 + x + 1 are (a – b), a and (a + b).

Now by using the relationship between the zeros of the quadratic polynomial we have:

Sum of zeros = -(coefficient of x)/(coefficient of x^2) = -1

a – b + a + a + b = -1

3a = 3

a = 1

Product of zeros = (constant term)/(coefficient of x^2) = -1

(a – b) (a) (a + b) = -1

(1 – b) (1) (1 + b) = – 1

1 – b^2 = – 1

b = ±√2


 

RS Aggarwal Solutions for Class 10 Maths Chapter 2 Polynomials

In this chapter students will study important concepts on polynomials as listed below:

  • Linear Polynomials
  • Quadratic Polynomials
  • Cubic Polynomials
  • Biquadratic Polynomials
  • Value of a Polynomials at a given point
  • Zeros of a Polynomials
  • Relation between the zeros and coefficients of a quadratic Polynomial

Key Features of RS Aggarwal Solutions for Class 10 Maths Chapter 2 Polynomials

1. RS Aggarwal Solutions provide fully resolved step by step solutions to all questions.

2. Polynomials study material prepared based on the latest CBSE syllabus by subject experts.

3. Easy for quick revision

4. Help to clear doubts and score good marks in the exams.

5. These solutions will help the students to strengthen their foundation on the topic.

 

BOOK

Free Class