Question

Find the real zeroes of the polynomial:

f(x)=2x(x+5)^2-8x

I know you can pull the factors from the leading coefficient and the constant:
constant = p= 5 factors 1, 5
leading coefficient= q= 2 factors 1, 2

so would my rational zeros be: +/-{1, 1/2, 5, 5/2} and at that point whats my next order of operation?

Answer 1

for the zeros of 2x(x+5)^2-8x
2x(x+5)^2-8x = 0
2x( (x+5)^2 - 4) = 0
x = 0 or (x+5)^2 = 4
x = 0 or x+5 = ±2

x = 0, -3, -7

proof:
http://www.wolframalpha.com/input/?i=solve+2x(x%2B5)%5E2-8x+%3D0

I have no idea where you got your answers from, but they are not the zeroes of
your given function.