Ex-2.2 Polynomials, Class 10, Maths RD Sharma Solutions PDF Download

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

(i) f(x) = 2x³ + x² – 5x + 2;1/2,1,−2

(ii) g(x) = x³ – 4x² + 5x – 2;2,1,1

Sol: (i) f(x) = 2x³ + x² – 5x + 2;1/2,1,−2

(a) By putting x =  1/2 in the above equation, we will get

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

=  f(1/2) = 2(1/8) + 1/4 – 5/2 + 2

=  f(1/2) = 1/4 + 1/4 – 5/2 + 2  = 0

(b) By putting x = 1 in the above equation, we will get

f(1) = 2(1)³ + (1)² – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

(c) By putting x = -2 in the above equation, we will get

f(−2) = 2(−2)³ + (−2)² – 5(−2) + 2

= -16 + 4 + 10 + 2 = -16 + 16 = 0

Now,

Sum of zeroes =  α + β + γ  =  −b/a

⇒ 1/2 + 1−2 = −1/2

−1/2 = −1/2

Product of the zeroes =  αβ + βγ + αγ  =  c/a

1/2×1 + 1×(−2) + (−2)×1/2 = −5/2

1/2 – 2 – 1 = −5/2

−5/2 = −5/2

Hence, verified.

(ii) g(x) = x³ – 4x² + 5x – 2;2,1,1

(a) By putting x = 2 in the given equation, we will get

g(2) = (2)³ – 4(2)² + 5(2) – 2

= 8 – 16 + 10 – 2 = 18 – 18 = 0

(b) By putting x = 1 in the given equation, we will get

g(1) = (1)³ – 4(1)² + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

Now,

Sum of zeroes =  α + β + γ  =  −b/a

⇒ 2 + 1 + 1 = −(−4)

4 = 4

Product of the zeroes =  αβ + βγ + αγ  =  c/a

2×1 + 1×1 + 1×2 = 5

2 + 1 + 2 = 5

5 = 5

αβγ = – (−2)

2×1×1  = 2

2 = 2

Hence, verified.

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

Sol: Any cubic polynomial is of the form ax³ + bx² + cx + d :

= x³  – (sum of the zeroes) x²  + (sum of the products of its zeroes) x – (product of the zeroes)

=  x³ – 3x² + (−1)x + (−3)

=  k[x³ – 3x² – x – 3]

k is any non-zero real numbers.

Q.3: If the zeroes of the polynomial f(x) = 2x³ – 15x² + 37x – 30, find them.

Sol: Let, α = a – d,β = a and γ = a + d be the zeroes of the polynomial.

f(x) = 2x³ – 15x² + 37x – 30

α + β + γ = – (−15/2) = 15/2

αβγ = – (−30/2) = 15

a – d + a + a + d =  15/2 and a(a – d)(a + d) = 15

So, 3a =  15/2

a =  5/2

And, a(a² + d²) = 15

d² = 1/4

d = 1/2

Therefore, α = 5/2 – 1/2 = 4/2 = 2

β = 5/2

γ = 5/2 + 1/2  = 3

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

Sol: f(x) = x³ + 3px² + 3qx + r

Let, a – d, a, a + d be the zeroes of the polynomial.

Then,

The sum of zeroes =  −b/a

a + a – d + a + d = -3p

3a = -3p

a = -p

Since, a is the zero of the polynomial f(x),

Therefore, f(a) = 0

f(a) = a³ + 3pa² + 3qa + r  = 0

Therefore, f(a) = 0

=  ⇒ a³ + 3pa² + 3qa + r = 0

=  ⇒ (−p)³ + 3p(−p)² + 3q(−p) + r = 0

=  −p³ + 3p³ – pq + r = 0

=  2p³ – pq + r = 0

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

Sol: f(x) = x³ + 3px² + 3qx + r

Let, a – d, a, a + d be the zeroes of the polynomial.

Then,

The sum of zeroes =  −b/a

a + a – d + a + d =  −3b/a

⇒ 3a = −3b/a

⇒ a = −3b/3a = −b/a

Since,f(a) = 0

⇒ a(a²) + 3b(a)² + 3c(a) + d = 0

2b³ – 3abc + a²d = 0

Q.6: If the zeroes of the polynomial f(x) = x³ – 12x² + 39x + k are in A.P., find the value of k.

Sol: 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

Now,

f(a) = 0

f(a) = a³ – 12a² + 39a + k

f(4) = 4³ – 12(4)² + 39(4) + k = 0

64 – 192 + 156 + k = 0

k = -28