(Q4)If the zeros of the polynomial x3-3x2 + x + 1 are a-b, a, a+b, find a and b.

Answer:

Given polynomial x3-3x2+x+1

Since,(a-b),a,(a+b) are three zeros of the polynomial x3-3x2+x+1

Therefore, sum of the zeros

= (a-b)+a+(a+b)=
-(-)
1
 =

So, 3a= a=

Sum of the products of its zeros taken two at a time

a(a-b)+a(a+b)+(a+b)(a-b)=

a2-ab+a2+ab+a2-b2=

3a2-b2=

So, 3(1)2-b2=

3-b2=

b2=

b= √

Here a= and b=√