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$$x^3+6 x^2+11x +6 = (x+1)(x+2)(x+3)$$

Looking for a very detailed description of how to factor this, and just factoring in general.

Additionally, what type of problem is this, so I can make better and more relevant searches for help on future questions. Is it a cubic trinomial?

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You know the Rational Root Theorem? –  J. M. Nov 6 '11 at 18:40
    
I'm reading about it right now en.wikipedia.org/wiki/Rational_root_theorem –  xsari3x Nov 6 '11 at 18:44

1 Answer 1

up vote 1 down vote accepted

The easiest way, but unfortunatelly only works some times is the following:

Step 1: Guess a root $a$ of $P(x)$. Step 2: Divide by $x-a$. Step 3: Factor the remaining quadratic.

The following usually helps:

Let $P(x)=a_3x^3+a_2x^2+a_1x+a_0$. If all coefficients are integers, and $r=\frac{p}{q}$ is a irreducible rational root, then $p|a_0$ and $q|a_3$.

So for your problem, the only possible rational roots are $\pm 1, \pm 2, \pm 3$ and $\pm 6$, either one works or you need to rely on the cubic formula (which you can find easely but it is hard to memorize)...

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ehm , how to rely on the cubic formula ?? Thanks in advance –  xsari3x Nov 6 '11 at 18:47
    
I think I got the idea –  xsari3x Nov 6 '11 at 19:00
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Any factoring you find in "precalculus" will not involve the cubic formula. –  GEdgar Nov 6 '11 at 21:07

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