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?
 A: It's not a special case of any general factorization rule (except perhaps the Rational Root Test).
However, one easy ad-hoc way to derive the factorization is to notice
$\rm\quad\quad\quad\  f(x) + 3\:(x-3)\ =\ x\: (x^2 - 9)  $  
Thus $\rm\ \ f(x)\ =\ (x-3)\ (x\: (x+3) - 3)$ 
A: One way to see that is that $x=3$ is a root of $x^3 - 12x  + 9$. So $(x-3)$ will be a factor (by the Factor Theorem).
Now you can try to get $x-3$ somehow.
One way you can do that is to rewrite
$$x^3-12x + 9 = x^3 - (3^3 - 3^3) - 12(x - 3+3)  + 9$$
$$ = x^3 - 27 + 27 - 12(x - 3) - 36 + 9 = x^3 - 27 - 12(x-3) + (9 -36 + 27)$$
$$ = (x-3)(x^2+3x+9) - 12(x-3) = (x-3)(x^2+3x-3) $$
Here we used the fact that $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$
In general, if $r$ is a root of $f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{0}$, then $f(x) - f(r) = f(x)$ gives us a way to factorize $f(x)$ as $(x-r)g(x)$.
$$f(x) = a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{0} - (a_{n}r^{n} + a_{n-1}r^{n-1} + \dots +a_{0})$$
$$ = a_{n}(x^n - r^n) + a_{n-1}(x^{n-1} - r^{n-1}) + \dots + a_{1}(x-r)$$
Just like $x^3 - a^2 = (x-a)(x^2 + ax + a^2)$ we have that
$$x^n - r^n = (x-r)(x^{n-1} + rx^{n-1} + \dots + r^{n-1})$$
and so
$$f(x) = (x-r) (a_{n}(x^{n-1} + rx^{n-2} + \dots + r^{n-1}) + a_{n-1}(x^{n-2} + \dots +r^{n-1}) + \dots + a_1)$$
Once we know a root, we can also try using Polynomial Long Division to get the other factor.
For cubics, the roots can be found without the need to guess. Check this out: Cardano's Method.
Does that help?
A: By the Rational Zero Theorem all the rational roots of $x^{3}-12x+9$ must have a numerator which is a factor of $9$ and a denominator which is a factor of $1$. Therefore they have to be of the form $\frac{9}{1}=9$ or $\frac{3}{1}=3$. Let $f(x)=x^{3}-12x+9$. Since $f(9)=630$ and $f(3)=0$, $3$ is a root of $f(x)$. So it can be factored as
$x^{3}-12x+9=(x-3)\left( ax^{2}+bx+c\right) =ax^{3}+\left( b-3a\right)
x^{2}+\left( c-3b\right) x-3c$ 
Comparing coefficients we get
$a=1,b-3=0\iff b=3,-3c=9\iff c=-3$.
Then
$x^{3}-12x+9=(x-3)\left( x^{2}+3x-3\right) $.
PS.  Or we could apply Ruffini's rule to find the coefficients of $ax^{2}+bx+c$.
PPS. As commented by user1827 "The rational zero theorem" also permits "$1/1=1$ and $-9,-3,-1$. But they are not zeros." 
Of course, since $f(3)=0$ we can factor $f(x)$ rightaway, without taking into consideration all the remaining possibilities.
A: 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}
A: While possibly better answers have already been shown, I could not resist mentioning the ready-made (closed form) formula for this case. 
In general to factor a Polynomial, you need to know at least one root (a value where the polynomial becomes zero). 
If you know that the roots are $r1, r2$ and $r3$, you can write your equation as $(x-r1)(x-r2)(x-r3)$.
It would be so nice if you can find the root by inspection, however this is not possible for all of us. Fortunately, there are ready-made-formulas. 
For polynomial of degrees two there is Quadratic Formula-Wiki Ref..
For degree three (as in your case), there is Closed-Form for Polynomial Degree 3-Cardano. This formula is not usually used maybe because it is difficult to apply and may result not-so-precise results (when used by machine) due to the square roots. but it gives the answer mechanically without requiring too much knowledge of algebra. 
The formula provides a value for each of the 3 roots of the cubic equation. 
In your case, if you apply the formula you get $\left(x-3\right)\left(x+3.79134\right)\left(x-0.79134\right)$. The roots are accurate to 3 decimal places. 
Other methods exist too, such as Newton's and Cardano's.
A: I can use Ruffini's theorem, so: $p=\pm 1,\pm 3, \pm 9$ (divisors of the term of degree $0$) and $q=\pm 1$  (divisor of the term of maximum degree). $\frac{p}{q}=\pm 1,\pm 3, \pm 9$. If I plug into $x^3-12x+9$, $3$, I obtain $P(3)=27-36+9=0$: $-3$ is a rational zero of the polynomial. Now, dividing by $(x-3)$, I obtain: $x^2+3x-3$. $P(x)$ can be factored as: $$P(x)=(x-3)\cdot (x^2+3x-3)$$
A: Now x^3 −12x + 9
= x^3 − 9x - 3x +9
= x(x^2 - 9) - 3(x - 3)
= x(x -3)(x - 9) - 3(x - 3)
= (x - 3)[(x(x + 3) - 3]
= (x - 3)(x^2 + 3x - 3)
