Question

if the zeroes of the polynomial f(x)= x3-3x2+x+1 are a-b,a,a+b​

Answer 1

Answer:

p(x) = x3 - 3x2 + x+1

sum of zeros = cofficient of x2 / coffiecient of x3

a-b+a+a+b =-(-3)/1

3a=3

a=1

product of zeros=constant term/cofficient of x3

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

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

1square -b square =-1

-b square =-2

b square=2

b=+√2

hence proved, a=1, b=+√2