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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)$2011-12-17

4 Answers 4

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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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    @Jordan ... the integer factors of $ab=12$ are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$.2011-12-18
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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 only have a couple weeks before I attempt calculus again not sure how much practice I can get in.2011-12-17
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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)