# 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?

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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.

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you meant $\frac{9}{1}=9$ instead of 1 (second sentence) –  Eugene Bulkin Sep 28 '10 at 21:41
Eugene Bulkin: Thanks! Corrected. –  Américo Tavares Sep 28 '10 at 22:24
The rational zero theorem would also permit 1/1=1 and -9, -3, -1. But they are not zeros. –  Ross Millikan Sep 28 '10 at 23:06
user1827: yes, the symmetric values of $9$ and $3$ are also allowed, as well as $\pm 1$. Answer updated (corrected). Thanks for pointing that out! –  Américo Tavares Sep 28 '10 at 23:19

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 $\rm\ \ f(x)\ =\ (x-3)\ (x\: (x+3) - 3)$

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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?

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en.wikipedia.org/wiki/Factor_theorem might also be useful. –  Hans Lundmark Sep 28 '10 at 20:12
@Hans: Right, I will edit that into the answer. Thanks. –  Aryabhata Sep 28 '10 at 20:13
Cardano is a bit of a sledgehammer here; the only other thing I'll note is that synthetic division may be more convenient than long division, depending on the user. –  Ｊ. Ｍ. Sep 28 '10 at 23:58
@J.M: Right, just wanted to mention Cardano though, not for this, but in general. –  Aryabhata Sep 29 '10 at 14:32