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

 Chapter Name RD Sharma Chapter 2 Polynomials Book Name RD Sharma Mathematics for Class 10 Other Exercises Exercise 2.1Exercise 2.3 Related Study NCERT Solutions for Class 10 Maths

### Exercise 2.2 Solutions

1.  Verify that the numbers given alongside of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case:
(i) f(x) = 2x3 + x2  - 5x + 2; 1/2, 1 , -2
(ii) g(x) = x3 - 4x2 + 5x - 2; 2, 1, 1
Solution

(i) f(x) = 2x3 + x2 - 5x + 2
f(1/2) = 2(1/2)3 + (1/2)2 - 5(1/2) + 2
= 2/8 + 1/4 - 5/2 + 2 = -4/2 + 2 = 0
f(1) = 2(1)3 + (1)2 - 5(1) + 2
= 2 + 1 - 5+ 2 = 0
f(-2) = q(-2)3 + (-2)2 - 5(-2) + 2
= -16 + 4 + 10 + 2
= -16 + 16 = 0
= Î± + ÃŸ +  Î³ = -b/a
⇒ 1/2 + 1 - 2 = -1/2
⇒ 1/2 - 1 = -1/2
⇒ 1/2 = -1/2
Î±ÃŸ.ÃŸÎ³ +  Î³Î± = c/Î±
⇒ 1/2 ×1 + 1 × - 2 + -2 ×1/2 = -5/2
⇒ 1/2 - 2 - 1 = - 5/2
⇒ -5/2 = -5/2

(ii) g(x) = x3 - 4x2 + 5x - 2

g(2) = (2)3 - 4(2)2 + 5(2) - 2 = 8 - 16 + 10 - 2 = 18 - 18 = 0
g(1) = [1]3 - 4(1)2 + 5[1] - 2 = 1 - 4 + 5 - 2 = 6 - 6 = 0
Î± + ÃŸ + Î³ = -b/a(2) + 1 + 1 = -(-4) = 4 = 4
Î±ÃŸ + ÃŸÎ³ + Î³Î± = c/a
⇒ 2×1 + 1×1 + 1×2 = 5
⇒ 2 + 1 + 2 = 5
⇒ 5 = 5
Î±ÃŸÎ³ = -(-2)
⇒ 2×1×1 = 2
⇒ 2= 2

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

Any cubic polynomial is of the form ax3 + bx2 + cx +d
= x3 - Sum of zeroes(x2) [product of zeroes] + sum of the products of its zeroes x - product of zeroes

= x3 - 2x2 + (3 - x)+3
= k[x3 - 3x2 - x - 3]
k is any non - zero real numbers

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

Let Î± = a -d, ÃŸ= a and Î³ = a + d be the zeroes of polynomial.
f(x) = 2x3 - 15x2 + 37x - 30
Î± + ÃŸ + Î³ = -(-15/2) = 15/2
Î±ÃŸÎ³ = -(-30/2) = 15
a - d + a + a +d = 15/2 and a(a - d)(a+ a) = 15
3a = 15/2, a = 5/2
a(a2 - d2) = 15
⇒ a2 - a2  = (15×2)/5
⇒ (5/2)2 - d2 = 6

⇒ (25-6)/4 = d2
d2 = 1/4

⇒ d = 1/2
∴ Î± = 5/2 - 1/2 = 4/2 = 2
ÃŸ = 5/2 = 5/2
Î³ = 5/2 + 1/2 = 3

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

f(x) = x3 + 3px2 + 3qx +q
Let a - d, a, a+d be the zeroes of the polynomial

The sum of zeroes  -b/a
a + a - d + a +d = b/a
3a = -3p
a = -p
Since a is the zero of the polynomial f(x) therefore f(a) = 0
⇒ [a]3 + 3pa2 + 3qa + r = 0
∴ f(a) = 0
⇒ [a]3 + 3pa2 + 3qa + r = 0
⇒ p3 + 3p(-p)2 + 3q(-p) + r = 0
⇒ -p3 + 3p2  - pq + r = 0
⇒ 2p3 - pq + r = 0

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

Let a-d, a, a+ d be the zeroes of the polynomial f(x)
The sum of zeroes

⇒ a -d + a + a +d = -3b/a
⇒ +3a = -3b/a

⇒ a  = -3b/(a×3)

⇒ a = -b/a
f(a)  = 0

⇒ a(a)2 + 3b(a)2 + 3c(a) + d = 0
= a(-b/a)3 + 3b2/ a2 - (3bc/a) +d = 0
⇒ 2b3/a2 - 3bc/a  + d = 0
⇒ (2b3 - 3abc + a2d)/a2  = 0
⇒ 2b3 - 3abc + a2d = 0

6. If the zeroes of the polynomial f(x) = x3 - 12x2 + 39x + k are in A.P., find the value of k.
Solution
f(x) = x3 - 12x2 + 39x - k
Let a -d, a, a + d  be the zeroes of the polynomial f(x)
The sum of the zeroes = 12
3a = 12
⇒ a = 4
f(a) = -a(x)3 - l2(4)2 + 39(4) + k = 0
⇒ 64 - 192 + 156 + k = 0
⇒ -28 = k
⇒ k = -28