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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
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up vote 8 down vote accepted

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
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@Jordan Concerning the rational root test. The independent term of the polynomial $P(x)=x^{3}-3x^{2}-4x+12$ is $12$ and the coefficient of $x^{3}$ is $1$. The divisors of $12$ are $1,2,3,4,6,12$. By the rational root theorem the possible rational (integer in this case) roots of $P(x)=0$ may only be these divisors and their symmetric ones. We can e.g. compute $P(1)=6\neq 0$ and then $P(2)=0 $. Since $2$ is a zero of $P(x)$ we can find $p,q$ such that $P(x)=x^{3}-3x^{2}-4x+12=(x-2)(x^{2}+px+q)$. –  Américo Tavares Dec 18 '11 at 0:23
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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 $3$.

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I would never notice that. –  user138246 Dec 17 '11 at 18:54
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Sure you would. With a lot more practice. ;-) –  Dimitrije Kostic Dec 17 '11 at 19:04
    
I only have a couple weeks before I attempt calculus again not sure how much practice I can get in. –  user138246 Dec 17 '11 at 19:55
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Notice that $$ x^3-3x^2-4x+12 = x^2(x - 3) - 4(x-3) = (x^2 - 4)(x-3) = (x-2)(x+2)(x-3). $$

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x^3-3x^2-4x+12

When you have four terms in a polynomial, first look to see if you can group them two at a time and pull things out. For this one you can.

(x^3-3x^2)-4(x-3) = x^2(x-3)-4(x-3) = (x-3)(x^2-4) = (x-3)(x+2)(x-2)

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