# Factoring a Polynomial Having No Rational Roots

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.

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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. – Qiaochu Yuan Aug 12 '12 at 22:45
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 $\mathbb{Q}$. – Qiaochu Yuan Aug 12 '12 at 22:50
@QiaochuYuan You are correct – Sven Aug 12 '12 at 22:53
@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. – tomasz Aug 13 '12 at 1:51
For that matter, $$x^4 + 4 = (x^2 + 2 x + 2)(x^2 - 2 x + 2)$$ – Will Jagy Aug 13 '12 at 2:19

## 3 Answers

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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Two things you can use:

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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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