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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?

Rodrigo de Azevedo
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4 Answers4

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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.

Community
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Bill Dubuque
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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. –  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. –  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. –  Dec 18 '11 at 00:13
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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 00:23
  • @Jordan Note that $\rm:a,b:$ denote arbitrary integers above. The point was to generalize your example slightly to make the innate structure more evident. Here's a Factor Theorem link. These are topics that are normally taught in high school algebra courses. – Bill Dubuque Dec 18 '11 at 00:35
  • @Jordan ... the integer factors of $ab=12$ are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$. – Américo Tavares Dec 18 '11 at 00:35
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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 $3$.

Dimitrije Kostic
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Notice that $$ x^3-3x^2-4x+12 = x^2(x - 3) - 4(x-3) = (x^2 - 4)(x-3) = (x-2)(x+2)(x-3). $$

xen
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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)

King Squirrel
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