Factoring cubic 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?
 A: 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.
A: 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 $3$.
A: Notice that
$$
x^3-3x^2-4x+12 = x^2(x - 3) - 4(x-3) = (x^2 - 4)(x-3) = (x-2)(x+2)(x-3).
$$
A: 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)
