# Factoring 4 term polynomial

Trying to figure this one out but I see no logical approach to this at all.

$x^3-3x^2-4x+12$ I know that it will be 3 parts most likely and that each will start with x but beyond that I will just guess at evrything most likely. How do I factor something weird like this?

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$x^2(x-3)-4(x-3)$ –  pedja Dec 17 '11 at 18:24

HINT $\rm\quad f(x)\ =\ x^3 - a\ x^2 - b\ x + a\:b\: \ =\: \ x^2\ (x- a) - b\ (x - a)\ =\ \cdots$

Alternatively, by the Rational Root Test, the only possible rational roots are integer factors of $\rm\:a\:b\:.\:$ But clearly $\rm\ x = a\$ is a root since it makes the first and last pair of terms cancel out. Therefore, since $\rm\:f(a) = 0\:$ we deduce that $\rm\:f(x)\:$ has the factor $\rm\:x-a\:$ by the Factor Theorem.

For more efficient polynomial factorization algorithms see my post here and the following survey.

Kaltofen, E. Factorization of Polynomials, pp. 95-113 in:
Computer Algebra, B. Buchberger, R. Loos, G. Collins, editors, Vienna, Austria, 1982.

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This looks like a helpful comment but really it all goes over my head and I have no idea what is even being discussed. I am not familiar with much of the language being used. –  user138246 Dec 17 '11 at 22:05
@Jordan What is unfamiliar? What is your math background? –  Bill Dubuque Dec 17 '11 at 23:21
I have taken college algebra a couple times and calculus once, studying again for calc1. I am just having trouble following the discussion I have no idea where I am supposed to learn all this stuff I have never taken a class that taught this kind of material. –  user138246 Dec 17 '11 at 23:44
@Jordan But what precisely is it that you don't understand above. –  Bill Dubuque Dec 17 '11 at 23:53
The Rational Root test, integer factors of a b, x=a is a root, f(a)=0, f(x) has the factor x-a by the Factor Theorem, polynomial factorization algorithms, and then a lot of stuff from that link. –  user138246 Dec 18 '11 at 0:13
Notice that $$x^3-3x^2-4x+12 = x^2(x - 3) - 4(x-3) = (x^2 - 4)(x-3) = (x-2)(x+2)(x-3).$$
Notice that $x^3-3x^2-4x+12 = x^2(x-3) - 4(x-3) = (x^2-4)(x-3)$. So your roots are $x=-2,2,$ and $3$.