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I need to factor $P(x)$ in order to expand the fraction $\frac{1}{P(x)}$ in partial fractions. What I did I rewrote the original one as

$ P(x)=(b-\theta_1 x)(b-\theta_2x)(b_3 - \theta_3 x) $

then expanded it and equated coefficients at $x$ (e.g. for $x^3$ it would be $-\theta_1 \theta_2 \theta_3=b_3$ and so on). As a results algebra got really messy (e.g. see Wikipedia entry on roots of cubic function), so I wonder if there exists some easier method of doing it.

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    If the bit at Wikipedia goes too fast for you, you might want to do things the way Américo did [here](http://math.stackexchange.com/questions/60907/60913#60913).2011-11-09

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If the cubic doesn't factor over the integers, the answer is guaranteed to be messy. Let the roots of the cubic be $r,s,t$, so the cubic factors (over the complex numbers) as $b_3(x-r)(x-s)(x-t)$. Write

${1\over{\rm cubic}}={a\over x-r}+{b\over x-s}+{c\over x-t}$

multiply through by the cubic to get

$1=b_3(a(x-s)(x-t)+b(x-r)(x-t)+c(x-r)(x-s))$

evaluate succesively at $x=r$, $x=s$, and $x=t$ to get

$a={1\over b_3(r-s)(r-t)}$

etc.

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    @Tapu, you may be right. If OP wants to confirm that repeated roots were intended, I'll be happy to modify my answer.2012-01-11