RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.4

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) 2x³ + x² – 5x + 2, \(\frac{1}{2}\), 1, – 2
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Solution:
(i) Comparing the given polynomial with the polv-nomial ax³ + bx² + ex + d, 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
Therefore, \(\frac{1}{2}\), 1 and – 2 are the zeroes of the given polynomial 2x³ + x² – 5x + 2
i.e. α = \(\frac{1}{2}\), β = 1 and γ = – 2
Then,

Hence the sum and product by the zeroes \(\frac{1}{2}\). 1 and – 2 are the same as these.
∴ The relationship between the zeroes and the coefficients is correct.

(ii) Comparing the given polynomial with ax³ + bx² + cx + d, 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 arc the zeroes of the polynomial x³ – 4x² + 5x – 2
So, α = 2, β = 1 and γ = 1
Thus,
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 = \(\frac{-(-2)}{1}=\frac{-d}{a}\)
∴ The sum and product obtained by zeroes 2, 1. 1 are also these. Therefore the above relationship of the zeroes of the polynomial with its coefficient is correct.

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.
Solution:
Let a cubic polynomial be ax³ + bx² + cx + d and its zeroes are α, β and γ.
Then
α + β + γ = \(\frac{-(-2)}{1}=\frac{-b}{a}\)
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
Thus the polynomial is x³ – 2x² – 7x + 14

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b find a and b.
Solution:
∵ (a – b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + x + 1.
Therefore
(a – b) + a + (a – b) = \(\frac{-(-3)}{1}\) = 3
∵ α + β + γ = \(-\frac{b}{a}\)
⇒ 3a = 3
⇒ a = 1
(a – b) a + a (a + b) + (a + b) (a – b) = \(\frac{1}{1}\) = 1
∵ αβ + βγ + γα = \(\frac{c}{a}\)
⇒ a² – ab + a² + ab + a² – b² = 1
⇒ 3a² – b² = 1
⇒ 3(1)² – b² = 1 [∵ a = 1]
⇒ 3 – b² = 1
⇒ b² = 2
⇒ 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:
∵ Two zeroes of the polynomial p(x) = x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt{3}\)
Therefore, x = 2 ± \(\sqrt{3}\)
⇒ x – 2 = ± \(\sqrt{3}\)
Squaring both sides.
x² – 4x + 4 = 3
⇒ x² – 4x + 1 = 0
Now let us divide the polynomial p(x) by x² – 4x + 1
so that other zeroes may be obtained.

∴ 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)
⇒ (x + 5) and (x – 7) will be the other factors. Hence – 5 and 7 will be the other zeroes.

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:
Dividing the polynomial x⁴ – 6x³ + 16x² – 25x + 10 by the polynomial x² – 2 x + k

∴ Remaidner = (2k – 9) x – (8 – k) k + 10
But remainder = x + a
Therefore, Comparing the coefficients
2k – 9 = 1
⇒ 2k= 10
⇒ k = 5
and – (8 – k) k + 10 = a
⇒ a = – (8 – 5) 5 + 10
= – 3 × 5 + 10
= – 15 + 10 = – 5
Hence, k = 5 and a = – 5