# Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.4

Gujarat Board Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.4

## 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:

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)

⇒ 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 α = 2

β = 1

and γ = 1

Then, we have

α + β + γ = 2 + 1 + 1 = 4

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:

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 α = a – b

β = a

γ = a + b

Then, we have

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

⇒ 2k = 10 ⇒ k = 5

and -k(8 – k) + 10 = a

⇒ -5(8 – 5) + 10 = a

⇒ – 5 = a

⇒ a = -5

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