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

## EXERCISE 2.4

### (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.