Question

What are all the possible rational zeros for `f(x)=5x^3-31x^2+31x-5` and how do you find all zeros?

Answer 1

`1/5, 1 and 5.`

Explanation:

graph{y-5x^3+31x^2-31x+5=0 [-10, 10, -5, 5]} The sum of the coefficients is 0. So, 1 is a zero for f.

Now, `f(x)=(x-1)(5x^2-26x+5)`.

The zeros of the quadratic factor are

`(26+-sqrt(26^2-100))/10=(26+-24)/10=5 and 1/5`.

The x-intercepts in the graph give the zeros of f.