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• RD Sharma Class 10 Solutions

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

### (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

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