Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

4 Answers 4

up vote 6 down vote accepted

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.

share|improve this answer
    
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
2  
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 $\rm\ \ f(x)\ =\ (x-3)\ (x\: (x+3) - 3)$

share|improve this answer

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?

share|improve this answer
1  
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

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:

$x^3 - 12x + 9 \\= x^3 - 3x^2 + 3x^2 - 9x - 3x + 9 \\= x^2(x - 3) + 3x(x - 3) - 3(x - 3) \\= (x - 3)(x^2 + 3x - 3)$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.