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

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:

1. 2x³ + x² – 5x + 2; \(\frac{1}{2}\), 1, – 2
2. x³ – 4x² + 5x – 2; 2, 1, 1

Solution:
1. Comparing the given polynomial with ax³ + bx² + 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 2x³ + x² – 5x + 2.
So, α = \(\frac{1}{2}\), β = 1 and γ = – 2.

2. Comparing the given polynomial with ax³ + bx² + 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 x³ – 4x² + 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 ax³ + bx² + 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 x³ – 2x² – 7x + 14.

Question 3.
If the zeroes of the polynomial x³ – 3x² + 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 x³ – 3x² + 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 a² – ab + a² + ab + a² – b² = 1
or 3a² – b² = 1
So, 3(1)² – b² = 1 [∵ a = 1]
or 3 – b² = 1
or b² = 2 or b = ± \(\sqrt{2}\)
Hence, a = 1 and b = ± \(\sqrt{2}\).

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt{3}\), find other zeroes.
Solution:
We have:
2 ± \(\sqrt{3}\) are two zeroes of the polynomial p(x) = x⁴ – 6x³ – 26x² + 138x – 35
Let x = 2 ± \(\sqrt{3}\). So, x – 2 = ± \(\sqrt{3}\)
Squaring, we get
x² – 4x + 4 = 3, i.e; x² – 4x + 1 = 0
Let us divide p(x) by x² – 4x + 1 to obtain other zeroes.

∴ p(x) = x⁴ – 6x³ – 26x² + 138x – 35
= (x² – 4x + 1)(x² – 2x – 35)
= (x² – 4x + 1)(x² – 7x + 5x – 35)
= (x² – 4x + 1)[x(x – 7) + 5(x – 7)]
= (x² – 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 x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
Let us divide x⁴ – 6x³ + 16x² – 25x + 10 by x² – 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.