# Math Help - solving for x intercept

1. ## solving for x intercept

$y=2-15x+9x^2-x^3$
so $0=2-15x+9x^2-x^3$
I don't see anyway to factor this out, does this mean that there is no x intercept?

2. By graphing it, or otherwise, you can figure out that x=2 is a root. So factor out (x-2) to get $(x-2)(x^2-7x+1)=0$. Now can you factor the quadratic?

3. Originally Posted by dorkymichelle
$y=2-15x+9x^2-x^3$
so $0=2-15x+9x^2-x^3$
I don't see anyway to factor this out, does this mean that there is no x intercept?
all cubic polynomials have at least one real zero.

using the rational root theorem reveals that x = 2 is a real root of this cubic.

there may be more ... what can you do now to find them if they exist?

4. yes thanks, I can find the roots with the quadratic formula, I didnt think of factoring out the (x-2)