find the real zeros of the following function f(x)=x^3-5x^2+5x-1

Question

I  need to figure this out for Alegbra II.

Answer 1

By inspection we can see that x=1 is a zero: 1-5+5-1=0. Using synthetic or algebraic division we can divide (x-1) into the polynomial to get x^2-4x+1. Use the quadratic formula to get the two zeroes: x=(4+sqrt(16-4))/2=(4+sqrt(12))/2=(4+2sqrt(3))/2=2+sqrt(3)=3.732 and 0.268. So we have the three zeroes: 0.268, 1, 3.732 (approx. decimals).