<s>[INST] Factoring a Cubic Polynomial I've been trying to understand how  
${x^3-12x+9}$ 
factors to 
$(x-3) (x^2+3 x-3)$
What factoring rule does this follow? The net result seems to be similar to what is attained through the sum/difference of cubes factoring pattern, but the signs are different.
Additionally, what type of problem is this, so I can make better and more relevant searches for help on future questions. Is it a cubic trinomial?
 [/INST] In order to factor any cubic, you must find at least one root. You acknolwedged that $3$ is a root, thus $x=3$ and $x-3=0$.  And since $x-3$ is a factor of $x^3-12x+9$, split the polynomial in accordance with $x-3$ and factor as follows:
\begin{align}
x^3-12x+9 &= x^3-3x^2+3x^2-9x-3x+9\\
&=x^2(x-3)+3x(x-3)-3(x-3)\\
&=(x-3)(x^2+3x-3)
\end{align}
</s>