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## Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Ex 9.4

Question 1.

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x^{3} + x^{2} – 5x + 2; \(\frac{1}{2}\), 1, – 2

(ii) x^{3} – 4x^{2} + 5x – 2; 2, 1, 1

Answer:

Let P(x) = 2x^{3} + x^{2} – 5x + 2 on comparing with general polynomial P(x) = ax^{3} + bx^{2} + cx + d we get

a = 2, b = 1, c = – 5 and d = 2

Given zeroes \(\frac{1}{2}\), 1, – 2

P(1) = 2 (1)^{3} + (1)^{2} – 5(1) + 2

= 2 × 1 + 1 – 5 + 2

= 2 + 1 – 5 + 2 = 5 – 5 = 0

P(1) = 0

P(- 2) = 2(- 2)^{3} + (-2)^{2} – 5(- 2) + 2

= 2 × – 8 + 4 + 10 + 2

= – 16 + 16

P(- 2) = 0

Hence 1/2, 1 & -2 are the zeroes of the given cubic polynomial

Again consider α = 1/2, β = 1 & γ = – 2

ii) Let P(x) = x^{3} – 4x^{2} + 5x – 2 on comparing with general polynomial

P(x) = ax^{3} + bx^{2} + cx + d

we get a = 1, b = – 4, c = 5 & d = 2.

Given zeroes 2, 1, 1

P(2) = (2)^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 4 × 4 + 10 – 2

= 8 – 16 + 10 – 2

P(2) = 18 – 18 = 0

P(2) = 0

P(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4 + 5 – 2

P(1) = 6 – 6 = 0

P(1) = 0

Hence, 2, 1 and 1 are the zeroes of the given cubic polynomial.

Again, consider α = 2, β = 1 & γ = 1

α + β + γ = 2 + 1 + 1 = 4

and α + β + γ = \(\frac{-b}{a}=\frac{-(-4)}{1}\) = 4

αβ + βγ + γα = 2(1) + (1) (1) + (1) × (2)

= 2 + 1 + 2 = 5

αβ + βγ + γα = \(\frac{c}{a}=\frac{5}{l}\) = 5

αβγ = 2 × 1 × 1 = 2

αβγ = \(\frac{-d}{a}=\frac{-(-2)}{1}=\frac{2}{1}\) = 2

Question 2.

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.

Answer:

Let the cubic polynomial be P(x) = ax^{3} + bx^{2} + cx + d

Then sum of zeroes = \(\frac{-b}{a}\) = 2

Sum of the product of zeroes taken two at a time = \(\frac{c}{a}\) = – 7

Product of zeroes = \(\frac{-d}{a}\) = – 14

Question 3.

If the zeroes of the polynomial x^{3} – 3x^{2} + x + 1 are a – b, a, a + b, find a and b.

Answer:

The given Polynomial is x^{3} – 3x^{2} + x + 1

comparing with Ax^{3} + Bx^{2} + Cx + D,

A = 1, B = – 3, C = 1 & D = 1

Let α = a – b, β = a & γ = a + b

α + β + γ = \(\frac{B}{A}=\frac{-(-3)}{1}\) = 3

a – b + a + a + b = 3

3a = 3

a = \(\frac{3}{3}\) = 1

a = 1

αβγ = \(\frac{-D}{A}\) = – 1

(a – b) a (a + b) = – 1 Put a = 1

(1 – b) 1 (1 + b) = – 1

1 – b^{2} = – 1

1 + 1 = b^{2}

b^{2} = 2

b = ±\(\sqrt{2}\)

∴ a = 1 & b = ±\(\sqrt{2}\)

Question 4.

If two zeroes of the polynomial x^{4} – 6x^{3} – 26x^{2} + 138x – 35 are 2 ±\(\sqrt{3}\). find other zeroes.

Answer:

Two zeroes of the given polynomial x^{4} – 6x^{3} – 26x^{2} + 138x – 35 are 2 ± \(\sqrt{3}\).

∴ x = 2 ± \(\sqrt{3}\).

Therefore [(x – (2 + \(\sqrt{3}\))] [x – (2 – \(\sqrt{3}\))]

= (x – 2 – \(\sqrt{3}\)) (x – 2 + \(\sqrt{3}\))

= [(x – 2) – \(\sqrt{3}\)][(x – 2) + \(\sqrt{3}\))]

= (x – 2)^{2} – (\(\sqrt{3}\))^{2}

= x^{2} – 4x + 4 – 3 = x^{2} – 4x + 1

∴ It is factor of Polynomial apply division algorithm.

Other two zeroes are given by the Quotient

x^{2} – 2x – 35 = 0

x^{2} – 7x + 5x – 35 = 0

x(x – 7) + 5(x – 7) = 0

(x – 7)(x + 5) = 0

x – 7 = 0 (or) x + 5 = 0

x = 7 (or) x = – 5

∴ x = 7, – 5

Question 5.

If the polynomial x^{4} – 6x^{3} + 16x^{2} – 25x + 10 is divided by another polynomial x^{2} – 2x + k, the remainder comes out to be x + a, find k and a.

Answer:

Let us apply the division algorithm to the given Polynomial

But remainder is given to be x + a comparing coefficients of x in the remainder obtained and remainder given:

(2K – 9)x – K(8 – K)+ 10 = x + a

∴ 2K – 9 = 1

∴ 2K = 1 + 9

K = \(\frac{10}{2}\)

K = 5 and – K(8 – K) + 10 = a

Put K = 5

– 5(8 – 5) + 10 = a

– 5(3) + 10 = a

– 15 + 10 = a

a = – 5