Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.

Verify that the numbers given alongside of the cubic polynomials below are their zeros.

Also verify the relationship between the zeroes and the coefficients in each case:

- 2x
^{3}+ x^{2}– 5x + 2; \(\frac{1}{2}\), 1, – 2 - x
^{3}– 4x^{2}+ 5x – 2; 2, 1, 1

Solution:

1. Comparing the given polynomial with ax^{3} + bx^{2} + cx + d, we get

a = 2, b = 1, c = – 5 and d = 2.

p(1) = 2(1)^{3} + (1)^{2} – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

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

= 2(- 8) + 4 + 10 + 2

= – 16 + 16 = 0

∴ \(\frac{1}{2}\), 1 and – 2 are the zeroes of 2x^{3} + x^{2} – 5x + 2.

So, α = \(\frac{1}{2}\), β = 1 and γ = – 2.

2. Comparing the given polynomial with ax^{3} + bx^{2} + cx + d, we get

a = 1, b = – 4, c = 5 and d = – 2.

p(2) = (2)^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 16 + 10 – 2 = 0

p(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

∴ 2, 1 and 1 are the zeroes of x^{3} – 4x^{2} + 5x – 2.

So, α = 2, β = 1 and γ = 1.

Therefore, α + β + γ = 2 + 1 + 1 = 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\)

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

= 2 + 1 + 2 + 5 = \(\frac{5}{1}\) = \(\frac{c}{a}\)

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

Question 2.

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

Solution:

Let the cubic polynomial be ax^{3} + bx^{2} + cx + d, and its zeroes be α, β and γ.

Then, α + β + γ = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-b}{a}\)

αβ + βγ + γα = – 7 = \(\frac{-7}{1}\) = \(\frac{c}{a}\)

and αβγ = – 14 = \(\frac{-14}{1}\) = \(\frac{-d}{a}\)

If a = 1, then b = – 2, c = – 7 and d = 14.

So, one cubic polynomial which fits the given conditions x^{3} – 2x^{2} – 7x + 14.

Question 3.

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

Solution:

Since (a – b), a and (a + b) are the zeroes of the polynomial x^{3} – 3x^{2} + x + 1, therefore

So, (a – b) + a + (a + b) = \(\frac{-(-3)}{1}\) = 3

So, 3a = 3 or a = 1

(a – b)a + a(a + b) + (a + b)(a – b) = \(\frac{1}{1}\) = 1

or a^{2} – ab + a^{2} + ab + a^{2} – b^{2} = 1

or 3a^{2} – b^{2} = 1

So, 3(1)^{2} – b^{2} = 1 [∵ a = 1]

or 3 – b^{2} = 1

or b^{2} = 2 or b = ± \(\sqrt{2}\)

Hence, a = 1 and 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.

Solution:

We have:

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

Let x = 2 ± \(\sqrt{3}\). So, x – 2 = ± \(\sqrt{3}\)

Squaring, we get

x^{2} – 4x + 4 = 3, i.e; x^{2} – 4x + 1 = 0

Let us divide p(x) by x^{2} – 4x + 1 to obtain other zeroes.

∴ p(x) = x^{4} – 6x^{3} – 26x^{2} + 138x – 35

= (x^{2} – 4x + 1)(x^{2} – 2x – 35)

= (x^{2} – 4x + 1)(x^{2} – 7x + 5x – 35)

= (x^{2} – 4x + 1)[x(x – 7) + 5(x – 7)]

= (x^{2} – 4x + 1)(x + 5)(x – 7)

So, (x + 5) and (x – 7) are other factors of p(x).

∴ – 5 and 7 are other zeroes of the given polynomial.

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.

Solution:

Let us divide x^{4} – 6x^{3} + 16x^{2} – 25x + 10 by x^{2} – 2x + k.

∴ Remainder = (2k – 9)x – (8 – k)k + 10

But the remainder is given as x + a.

On comparing their coefficients, we have:

2k – 9 = 1 or 2k = 10 or k = 5

and – (8 – k)k + 10 = a

So, a = – (8 – 5)5 + 10

= – 3 × 5 + 10 = – 15 + 10 = – 5

Hence, k = 5 and a = – 5.