$\displaystyle y=2-15x+9x^2-x^3$
so $\displaystyle 0=2-15x+9x^2-x^3$
I don't see anyway to factor this out, does this mean that there is no x intercept?
By graphing it, or otherwise, you can figure out that x=2 is a root. So factor out (x-2) to get $\displaystyle (x-2)(x^2-7x+1)=0$. Now can you factor the quadratic?
$\displaystyle y=2-15x+9x^2-x^3$
so $\displaystyle 0=2-15x+9x^2-x^3$
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?