Chapter 2: Polynomials Exercise – 2.2

Question: 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

Solution: 

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

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

(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

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

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.

 

Question: 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.

Solution:

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.

 

Question: 3

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

Solution: 

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

f(x) = 2x³ – 15x² + 37x – 30 

And, a (a² + d²) = 15

 

Question: 4

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

Solution: 

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) = 0f(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

 

Question: 5

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

Solution:

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

Since, f(a) = 0

⇒ a(a²) + 3b(a)² + 3c(a) + d = 0

⇒ a(a²) + 3b(a)² + 3c(a) + d = 0 

 

Question: 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

Now,

f(a) = 0

f(a) = a³ – 12a² + 39a + k f(4) = 4³ – 12(4)² + 39(4) + k = 0

f(4) = 4³ –12(4)² + 39(4) + k = 0

64 – 192 + 156 + k = 0

k = – 28