posted by Math factoring on .
Can someone factor this for me?
i could only get up to:
That's as far as you can go using rational numbers
(I ran x^3-2x^2-4x-8 = 0 through my "cubic equation solver and got one messy real root and 2 complex roots)
would there be another way to find the x intercepts? Cause that's what i was trying to achieve
you could sketch it and see appr. where it crosses the x-axis.
let f(x) = x^3 - 2x^2 - 4x - 8
If there are any rational factors, then test for
f(x) = 0 using x = ±1, ±2,±4, and ±8
none work, so there are no rational factors.