Question

If the zeros of the polynomial x^3-3x^2+x+1 are (a-b), a, (a+b), find a and b.​

Answer 1

Answer:

The values of a and b are:

⇒a=1

⇒b=± √2

Step-by-step explanation:

a−b,a,a+b are zeroes of x^3-3x^2+x+1.

⇒a−b+a+a+b= sum of zeroes = 3

⇒3a=3

⇒a=1

W.k.t,

(a−b)(a)(a+b)= product of zeroes =−1

⇒(1−b)(1+b)=−1.

⇒1−b^2=-1. ( since (a-b)*(a+b)=a^2-b^2)

⇒b^2=2

⇒b=± √2