UP Board Solutions for Class 10 Maths Chapter 2 Exercise 2.4 Polynomials

EXERCISE 2.4

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) 2x2+ x2 – 5x + 2; , 1,-2

(ii) x3 – 4x2 + 5x – 2; 2, 1, 1

Sol. (i) Comparing the given polynomial with ax3+ bx2 + cx + d, we get
a = 2, b= 1, c = – 5 and d = 2.


(ii) 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, a = 2, = 1 and y = 1.

2.Find a cubic polynomial with the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.

Sol. Let the cubic polynomial be ax3+ bx2 + cx + d, and its zeroes be a, and y.
Then,


If a = 1, then b = – 2, c = – 7 and d = 14.
So, one cubic polynomial which fits the given conditions will be x3 – 2x2 – 7x + 14.

3.If the zeroes of the polynomial x3-3x2+ x + 1 are a – b, a, a + b, find a and b.
Sol. Since (a – b), a, (a + b) are the zeroes of the polynomial x3-3x2+ x + 1


p(x) = x4 – 6x3– 26x2 + 138x – 35
= (x2 – 4x + 1) (x2 – 2x – 35)
= (x2-4x+ 1)(x2 – 78 + 58 – 35)
= (x2 – 4x + 1) (x (x – 7) + 5(x – 7)]
= (x2– 4x + 1)(x + 5)(x- 7)
(x + 5) and (x – 7) are other factors of p(x)

=> 5 and 7 are other zeroes of the given polynomial.

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.

Sol. Let us divide x4– 6x3 + 16x2 – 25x + 10 by x2 – 2x + k
22 – 2x + k.

Leave a Comment