This exercise majorly focuses on verifying the zeros and the relationship between the zeros and coefficients of a cubic polynomial. The RD Sharma Solutions Class 10 prepared by experts at BYJU’S can help students understand these concepts clearly following the latest CBSE patterns. As a result, the students can achieve higher marks in their examinations. Students can download **RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2** PDF provided below for a detailed explanation of problems in this exercise.

## RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2 Download PDF

### Access RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

**1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeros and coefficients in each of the following cases:**

**(i) f(x) = 2x ^{3 }+ x^{2 }– 5x + 2; 1/2, 1, -2**

**Solution: **

Given, f(x) = 2x^{3 }+ x^{2 }– 5x + 2, where a= 2, b= 1, c= -5 and d= 2

For x = 1/2

f(1/2) = 2(1/2)^{3} + (1/2)^{2} – 5(1/2) + 2

= 1/4 + 1/4 – 5/2 + 2 = 0

⇒ f(1/2) = 0, hence x = 1/2 is a root of the given polynomial.

For x = 1

f(1) = 2(1)^{3} + (1)^{2} – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

⇒ f(1) = 0, hence x = 1 is also a root of the given polynomial.

For x = -2

f(-2) = 2(-2)^{3} + (-2)^{2} – 5(-2) + 2

= -16 + 4 + 10 + 2 = 0

⇒ f(-2) = 0, hence x = -2 is also a root of the given polynomial.

Now,

Sum of zeros = -b/a

1/2 + 1 – 2 = – (1)/2

-1/2 = -1/2

Sum of the products of the zeros taken two at a time = c/a

(1/2 x 1) + (1 x -2) + (1/2 x -2) = -5/ 2

1/2 – 2 + (-1) = -5/2

-5/2 = -5/2

Product of zeros = – d/a

1/2 x 1 x (– 2) = -(2)/2

-1 = -1

Hence, the relationship between the zeros and coefficients is verified.

**(ii) g(x) = x ^{3 }– 4x^{2 }+ 5x – 2; 2, 1, 1**

**Solution: **

Given, g(x) = x^{3 }– 4x^{2 }+ 5x – 2, where a= 1, b= -4, c= 5 and d= -2

For x = 2

g(2) = (2)^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 16 + 10 – 2 = 0

⇒ f(2) = 0, hence x = 2 is a root of the given polynomial.

For x = 1

g(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

⇒ g(1) = 0, hence x = 1 is also a root of the given polynomial.

Now,

Sum of zeros = -b/a

1 + 1 + 2 = – (-4)/1

4 = 4

Sum of the products of the zeros taken two at a time = c/a

(1 x 1) + (1 x 2) + (2 x 1) = 5/ 1

1 + 2 + 2 = 5

5 = 5

Product of zeros = – d/a

1 x 1 x 2 = -(-2)/1

2 = 2

Hence, the relationship between the zeros and coefficients is verified.

**2.** **Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and product of its zeros as 3, -1 and -3 respectively.**

**Solution: **

**Generally, **

A cubic polynomial say, f(x) is of the form ax^{3 }+ bx^{2 }+ cx + d.

And, can be shown w.r.t its relationship between roots as.

⇒ f(x) = k [x^{3} – (sum of roots)x^{2} + (sum of products of roots taken two at a time)x – (product of roots)]

Where, k is any non-zero real number.

Here,

f(x) = k [x^{3} – (3)x^{2} + (-1)x – (-3)]

∴ f(x) = k [x^{3} – 3x^{2} – x + 3)]

where, k is any non-zero real number.

**3.** **If the zeros of the polynomial f(x) = 2x ^{3} – 15x^{2} + 37x – 30 are in A.P., find them.**

**Solution: **

Let the zeros of the given polynomial be α, β and γ. (3 zeros as it’s a cubic polynomial)

**And given, the zeros are in A.P. **

**So, let’s consider the roots as **

α = a – d, β = a and γ = a +d

Where, a is the first term and d is the common difference.

From given f(x), a= 2, b= -15, c= 37 and d= 30

⇒ Sum of roots = α + β + γ = (a – d) + a + (a + d) = 3a = (-b/a) = -(-15/2) = 15/2

So, calculating for a, we get 3a = 15/2 ⇒ a = 5/2

⇒ Product of roots = (a – d) x (a) x (a + d) = a(a^{2} –d^{2}) = -d/a = -(30)/2 = 15

⇒ a(a^{2} –d^{2}) = 15

Substituting the value of a, we get

⇒ (5/2)[(5/2)^{2} –d^{2}] = 15

⇒ 5[(25/4) –d^{2}] = 30

⇒ (25/4) – d^{2} = 6

⇒ 25 – 4d^{2} = 24

⇒ 1 = 4d^{2}

∴ d = 1/2 or -1/2

Taking d = 1/2 and a = 5/2

We get,

the zeros as 2, 5/2 and 3

Taking d = -1/2 and a = 5/2

We get,

the zeros as 3, 5/2 and 2

Good