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:

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

Solution:

1. Let p(x) = 2x^{3} + x^{2} – 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 2x^{3} + x^{2} – 5x + 2.

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

a = 2

b = 1

c = -5

d = 2

Let Î± = \(\frac{1}{2}\)

Î² = 1

and Î³ = -2

Then, we have

2. Let p(x) = x^{3} – 4x^{2} + 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 x^{3} – 4x^{2} + 5x – 2. Hence (x – 2) (x – 1), i.e., x^{2} – 3x + 2 is a factors of the given polynomial. Now, we apply the division algorithm to the given polynomial and x^{2} – 3x + 2.

So, x^{3} – 4x^{2} + 5x – 2

= (x^{2} – 3x + 2) (x – 1)

â‡’ x^{3} – 4x^{2} + 5x – 2 = (x – 2) (x – 1) (x – 1)

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

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

a = 1

b = -4

c = 5

d = – 2

Let Î± = 2

Î² = 1

and Î³ = 1

Then, we have

Î± + Î² + Î³ = 2 + 1 + 1 = 4

= –\(\frac{-4}{1}\) = –\(\frac{b}{a}\)

Î±Î² + Î²Î³ + Î±Î³ = (2)(1) + (1)(1) + (1)(2)

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

Î±Î²Î³ = (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 ax^{3} + bx^{2} + cx + d and its zeroes be Î±, Î² and Î³.

Then, Î± + Î² + Î³ = 2= \(\frac{-b}{a}\)

Î±Î² + Î²Î³ + Î³Î± = -7 = \(\frac{c}{a}\)

Î±Î²Î³ = -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

x^{3} – 2x^{2} – 7x + 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.

Solution:

Then given polynomial is x^{3} – 3x^{2} + x + 1

Comparing with Ax^{3} + Bx^{2} + Cx + D, we get

A = 1

B = -3

C = 1

D = 1

Let Î± = a – b

Î² = a

Î³ = a + b

Then, we have

Î± + Î² + Î³ = \(\frac{-B}{A}=-\frac{(-3)}{1}=3\)

â‡’ a – b + a + a + b = 3

â‡’ 3a = 3

â‡’ a = 1

Î±Î²Î³ = \(\frac{-D}{A}\) = -1

â‡’ (a – b) a (a + b) = -1

â‡’ (1 – b) 1(1 + b) = -1

â‡’ 1 – b^{2} = -1

â‡’ b^{2} = 2

â‡’ b = Â± \(\sqrt { 2 } \)

Question 4.

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

Solution:

Since two zeroes of the given polynomial x^{4} – 6x^{3} – 26x^{2} + 138x – 35 are 2 Â± \(\sqrt { 3 } \), 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., x^{2} – 4x + 1

is a factor of the given polynomial. Now, we apply the division algorithm to the given polynomial and x^{2} – 4x + 1.

So, x^{4} – 6x^{3} – 26x^{2} + 138x – 35

= (x^{2} – 4x + 1) (x^{2} – 2x – 35)

= (x^{2} – 4x + 1) (x – 7) (x + 5)

Hence, the other two zeroes are 7 and -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.

Solution:

Let us apply the division algorithm to the give Polynomial x^{4} – 6x^{3} + 16x^{2} – 25x + 10 and another Polynomial x^{2} – 2x + k.

Remainder

= (2k – 9)x – k (8 – k) + 10

But the remainder is given to be x + a.

Therefore, 2k – 9 = 1

â‡’ 2k = 10 â‡’ k = 5

and -k(8 – k) + 10 = a

â‡’ -5(8 – 5) + 10 = a

â‡’ – 5 = a

â‡’ a = -5