Class 10 RD Sharma Solutions- Chapter 2 Polynomials - Exercise 2.2

In this article, we will explore the solutions to Exercise 2.2 from Chapter 2 of RD Sharma's Class 10 Mathematics textbook which focuses on the "Polynomials". This chapter is fundamental for understanding the basics of polynomials their properties and how to manipulate them. By solving Exercise 2.2 students will gain hands-on experience in applying polynomial concepts and enhance their problem-solving skills.

Polynomials

The Polynomials are mathematical expressions involving a sum of powers of a variable multiplied by the coefficients. They are categorized by their degree which is the highest power of the variable in the expression. For example: 3x²+2x−5 is a polynomial of degree 2. Understanding polynomials involves operations like addition, subtraction, multiplication, and factorization which are crucial for solving a variety of algebraic problems.

Q1. 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) = 2x³ + x² - 5x + 2; 1/2, 1, -2

Solution:

For the followings numbers to be the zeros of polynomial they must satisfy the following equation i.e. f(x)=0 so now checking:

When x = 1/2

 f(1/2) = 2(1/2)³ + (1/2)² - 5(1/2) + 2 

f(1/2) = 1/4 + 1/4 - 5/2 + 2 = 0 

f(1/2) = 0,    

Hence, x = 1/2 is a zero of the given polynomial.

When x = 1

f(1) = 2(1)³ + (1)² - 5(1) + 2

f(1) = 2 + 1 - 5 + 2 = 0 

f(1) = 0,  

Hence, x = 1 is also a zero of the given polynomial.

When x = -2 

f(-2) = 2(-2)³ + (-2)² - 5(-2) + 2 

f(-2) = -16 + 4 + 10 + 2 = 0 

f(-2) = 0,   

Hence, x = -2 is also a zero of the given polynomial.

We know that Sum of zeros = -b/a;

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

Product of zeros = -d/a;

Here sum=1/2 + 1 - 2 =-1/2 and -b/a=-1/2

Here sum of products=(1/2 * 1) + (1 * -2) + (1/2 * -2) =-5/2 and c/a=-5/2

Here product =1/2 x 1 x (- 2) = -1 and -d/a=-1

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

(ii) g(x) = x³ – 4x² + 5x – 2; 2, 1, 1

Solution: 

For the followings numbers to be the zeros of polynomial they must satisfy the following equation ie. g(x)=0 so now checking:

When x = 2 

g(2) = (2)³ - 4(2)² + 5(2) - 2 

g(2) = 8 - 16 + 10 - 2 = 0 

g(2) = 0, 

Hence, x = 2 is a zero of the given polynomial.

Now we have two same roots, so we will check only once.

When x = 1 

g(1) = (1)³ - 4(1)² + 5(1) - 2 

g(1) = 1 - 4 + 5 - 2 = 0 

g(1) = 0, 

Hence, x = 1 is also a zero of the given polynomial.

We know that Sum of zeros = -b/a;

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

Product of zeros = - d/a;

Here sum=2+1+1= 4 and -b/a=4 

Here sum of products=(1 * 1) + (1 * 2) + (2 * 1) =5 and c/a =5

Here product = 2*1*1=2 and -d/a=2

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

Question 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: 

A cubic polynomial say, f(x) is of the form ax³ + bx² + cx + d.

We know that f(x) = k [x³ - (sum of roots)x² + (sum of products of roots taken two at a time)x -(product of roots)]

Sum of roots = 3; 

Sum of products of roots taken two at a time=-1; 

Product of roots=-3

f(x) = k [x³ - (3)x² + (-1)x - (-3)]

∴ f(x) = k [x³ - 3x² - x + 3)]

Required polynomial f(x) = k [x³ - (3)x² + (-1)x - (-3)]

∴ f(x) = k[x³ - 3x² - x + 3)]

Question 3. If the zeros of the polynomial f(x) = 2x³ – 15x² + 37x – 30 are in A.P., find them.

Solution:

 Let the roots be 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 i.e. a = 5/2

=> Product of roots = (a – d) x (a) x (a + d) = a(a2 –d2) = -d/a = -(30)/2 = 15 i.e. a(a2 –d2) = 15

Substituting 'a' we get ∴ d = 1/2 or -1/2

When d=1/2

Roots are α = 5/2-1/2, β = 5/2 and γ = 5/2 +1/2 i.e. α = 2, β = 2.5 and γ = 3

When d=-1/2

Roots are α = 5/2-(-1/2), β = 5/2 and γ = 5/2 +(-1/2) i.e. α = 3, β = 2.5 and γ = 2

Question 4. Find the condition that the zeros of polynomial f(x) = x³+3px²+3qx+r may be in A.P.

Solution:

 Let the roots be α = A - D, β = A and γ = A +D, Where, A is the first term and D is the common difference.

sum = A-D+A+A+D = -b/a i.e. 3A=-3b so, A=-b

Since β = A is a root so f(A)=0;

a³+3pa²+3qa+r=0

Now put a=-p; so, we get

2p²-3pq+r=0 is the required condition.

Question 5. If the zeros of the polynomial f(x) = ax³+3bx²+3cx+d are in A.P. Prove that 2b³-3abc+a²d = 0. 

Solution:

Let the roots be α = A - D, β = A and γ = A +D, Where, A is the first term and D is the common difference.

sum = A-D+A+A+D = -b/a i.e. 3A=-3b/a so, A=-b/a

Now f(A)=0 so,

f(x)=aA³+3bA²+3cA+d   

Now put A=-b/a; So, we get: 

2b³-3abc+a²d =0

Hence, proved.

Question 6. If the zeros of the polynomial f(x) = x³-12x²+39x+k are in A.P, find the value of k.

Solution:

Let the roots be α = A - D, β = A and γ = A +D, Where, A is the first term and D is the common difference.

sum = A-D+A+A+D = -b/a i.e., 3A=12 so, A=4

f(β)=0 i.e. f(A)=0

(4)³-12(4)²+39(4)+k=0

64-192+156+k=0

28+k=0

k=-28

Read More:

• What are Polynomials?
• Polynomial Formula

Summary

Exercise 2.2 of RD Sharma's Class 10 Mathematics textbook focuses on the division algorithm for polynomials. This section covers how to divide polynomials using long division and synthetic division methods. Students learn to find quotients and remainders when dividing polynomials, and understand the relationship between the degree of the dividend, divisor, quotient, and remainder. The exercise also includes problems on the factor theorem and applications of polynomial division in solving various mathematical problems.