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For example: $$ x^3-6x^2+3x-10 $$

The rational roots test tells me that possible roots are $\pm\ 10, 5, 2, 1$. However, none of these roots will divide the polynomial into a more workable nominal.

How can I efficiently determine how to factor this without resources such as Wolfram|Alpha?

Thank you.

  • 1
    A cubic polynomial can only factor into the product of a linear and a quadratic factor (which itself might factor further). So if it has no linear factors, it's already irreducible.2012-08-12
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    @QiaochuYuan It has a linear factor; it just doesn't have a *rational* root.2012-08-12
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    I am assuming (based on the fact that the OP said "possible roots" and not "possible rational roots") that the OP wants to factor over $\mathbb{Q}$.2012-08-12
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    @QiaochuYuan You are correct2012-08-12
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    @Sven In that case, none of the tools mentioned below will give you a factorisation over $\mathbb Q$, since you have proven yourself that **there is no rational root**2012-08-13
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    To rephrase some of the answers below, factoring over $\Bbb{R}$ amounts to finding real roots of the polynomial.2012-08-13
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    @JenniferDylan: No, it does not. $x^4+1$ has no real roots, but it does factor (over $\mathbf R$) into $(x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x+1)$. It does if degree is at most three.2012-08-13
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    For that matter, $$ x^4 + 4 = (x^2 + 2 x + 2)(x^2 - 2 x + 2) $$2012-08-13
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    @tomasz Oh my bad, I meant factoring into linear factors.2012-08-13

3 Answers 3

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Turgeon already mentioned Cardano; here's how to apply it to your polynomial.

The first thing you need to do is to depress your cubic. "Depression" is the generalization of the usual "completing the square" done on quadratics. In particular, we make a variable substitution that removes the quadratic term.

In this case, from Vieta's formulas, we know that the mean of the roots is $6/3=2$, so we let $x=u+2$. This yields the polynomial $u^3-9u-20$.

Now we come to Cardano's piece of trickery. Cardano assumes that a root $u$ of $u^3-9u-20$ can be written in the form $u=r+s$. If we make that substitution, we obtain

$$(r+s)^3-9(r+s)-20=r^3+s^3+3(r+s)(rs-3)-20$$

From this, we obtain the system of equations

$$\begin{align*} r^3+s^3&=20\\ r^3 s^3&=27 \end{align*}$$

Using Vieta's formulas again, we obtain the "quadratic" equation

$$(r^3)^2-20(r^3)+27=0$$

You should now be able to obtain $r$ and $s$. You already know that $x=r+s+2$ is one root of your cubic; divide that out of the original cubic, and solve the remaining quadratic. Done.

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By the intermediate value theorem, there's a root between $5$ and $6$. You can use Newton's method to quickly find a close approximation to it. Call that number $a$. Then $x-a$ is a factor. Use long division to find the quadratic factor. The arithmetic may get a bit messy, but after that you're just solving a quadratic equation.

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    How did you figure that it has a root between 5 and 6?2017-07-26
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    @JozemiteApps : As I said: "by the intermediate value theorem". The value of the polynomial is negative when $x=5$ and is positive when $x=6,$ and polynomial functions are continuous.2017-07-26
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Two things you can use: