GSEB Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.4

Gujarat Board GSEB Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.4 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 2 Polynomials Ex 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:

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

Solution:
1. Let p(x) = 2x³ + x² – 5x + 2
Then, we have
\(p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{2}\right)^{2}-5\left(\frac{1}{2}\right)+2\)
= \(\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=0\)
p(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
and, p(-2) = 2(-2)³ + (-2)² – 5(-2) + 2
= -16 + 4 + 10 + 2 = 0
Therefore, \(\frac{1}{2}\), 1 and -2 are the zeroes of 2x³ + x² – 5x + 2.
Comparing the given polynomial with ax³ + bx² + cx + d, we get
a = 2
b = 1
c = -5
d = 2
Let α = \(\frac{1}{2}\)
β = 1
and γ = -2
Then, we have

2. Let p(x) = x³ – 4x² + 5x – 2
Then, we have
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
Therefore, 2 and 1 are the two zeroes of x³ – 4x² + 5x – 2. Hence (x – 2) (x – 1), i.e., x² – 3x + 2 is a factors of the given polynomial. Now, we apply the division algorithm to the given polynomial and x² – 3x + 2.

So, x³ – 4x² + 5x – 2
= (x² – 3x + 2) (x – 1)
⇒ x³ – 4x² + 5x – 2 = (x – 2) (x – 1) (x – 1)
Hence, 2, 1 and 1 are the zeros of x³ – 4x² + 5x – 2.
Comparing the given polynomial with ax³ + bx² + cx + d, we get
a = 1
b = -4
c = 5
d = – 2
Let α = 2
β = 1
and γ = 1
Then, we have
α + β + γ = 2 + 1 + 1 = 4
= –\(\frac{-4}{1}\) = –\(\frac{b}{a}\)
αβ + βγ + αγ = (2)(1) + (1)(1) + (1)(2)
= 5 = \(\frac{5}{1}\) = \(\frac{c}{a}\)
αβγ = (2) (1) (1) = 2
= –\(\frac{-2}{1}\) = \(\frac{-d}{a}\)

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeros 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= \(\frac{-b}{a}\)
αβ + βγ + γα = -7 = \(\frac{c}{a}\)
αβγ = -14 = \(\frac{-d}{a}\)
If a = 1, then b = -2, c = -7 and d = 14.
Hence, one cubic polynomial which fits the given conditions 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:
Then given polynomial is x³ – 3x² + x + 1
Comparing with Ax³ + Bx² + Cx + D, we get
A = 1
B = -3
C = 1
D = 1
Let α = a – b
β = a
γ = a + b
Then, we have
α + β + γ = \(\frac{-B}{A}=-\frac{(-3)}{1}=3\)
⇒ a – b + a + a + b = 3
⇒ 3a = 3
⇒ a = 1
αβγ = \(\frac{-D}{A}\) = -1
⇒ (a – b) a (a + b) = -1
⇒ (1 – b) 1(1 + b) = -1
⇒ 1 – b² = -1
⇒ b² = 2
⇒ b = ± \(\sqrt { 2 } \)

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt { 3 } \), find the other zeroes.
Solution:
Since two zeroes of the given polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt { 3 } \), therefore,
(x – (2 + \(\sqrt { 3 } \))) (x – (2 – \(\sqrt { 3 } \))),
i.e., ((x – 2) – \(\sqrt { 3 } \)) (x – 2) + \(\sqrt { 3 } \))
i.e., (x – 2)² – (\(\sqrt { 3 } \))², i.e., x² – 4x + 1
is a factor of the given polynomial. Now, we apply the division algorithm to the given polynomial and x² – 4x + 1.

So, x⁴ – 6x³ – 26x² + 138x – 35
= (x² – 4x + 1) (x² – 2x – 35)
= (x² – 4x + 1) (x – 7) (x + 5)
Hence, the other two zeroes are 7 and -5.

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 apply the division algorithm to the give Polynomial x⁴ – 6x³ + 16x² – 25x + 10 and another Polynomial x² – 2x + k.

Remainder
= (2k – 9)x – k (8 – k) + 10
But the remainder is given to be x + a.
Therefore, 2k – 9 = 1
⇒ 2k = 10 ⇒ k = 5
and -k(8 – k) + 10 = a
⇒ -5(8 – 5) + 10 = a
⇒ – 5 = a
⇒ a = -5