# Factorization of a polynomial

I need to find the roots of this polynomial $$2x^2-x^4-x=0.$$ I know that it is necessary the factorization to obtain $$-x(x-1)(x^2+x-1)=0.$$ I asked to factorize my polynomial to Mathematica. The last step to find the roots is to split the polynomial in three equations.

What kind of method of factorization could I use to do things like this? I also know that there are a lot of methods of factorization; could you suggest me the most useful, simplest to start with?

Thank you.

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easy to see that $x=0$ and $x=1$ are the roots of the given equation. just factor it out. – Santosh Linkha Jan 30 '14 at 17:34
With small coefficients and only a couple terms, trial and error using synthetic division would definitely be a good place to start. – abiessu Jan 30 '14 at 17:34
Well, first of all, $x=0$ is obviously a root. Thus $2x^2- x^4 - x = 0 \Longrightarrow x(2x - x^3 -1) = 0$. Then it is easy to spot that $x=1$ is a root. Then, by long-division you can find the last factorization. – Nigel Overmars Jan 30 '14 at 17:35
The Rational Root Theorem is often useful in school exercises. – André Nicolas Jan 30 '14 at 17:36
@Santosh There are two more roots anyway $\frac12(-1\pm\sqrt5)$ – Charlie Jan 30 '14 at 17:39

You have already solved most of this by yourself. If you have a high-degree polynomial (say $\deg P > 2$), try to find some roots by "guessing", for example $x=0$ and $x=1$ in your polynomial. Finding a root, to get another root, you do polynomial (long) division: $$p(x) = 2x^2 - x^4 - x \stackrel!= 0$$ $x=0$ was found, so get $$p_1(x) := \frac{p(x)}{x} = 2x - x^3-1\stackrel!=0$$ Guessing another root, $x=1$, you get $$p_2(x) := \frac{p_1(x)}{x-1} = -x^2 -x +1 \stackrel!=0$$ From there ($\deg p_2 \le 2$) we can continue by completing the square: $$-(x^2 + x + \frac14 - \frac54) \stackrel!=0$$ Thus $$(x+\frac12)^2 \stackrel!= \frac54$$ $$\Rightarrow x \stackrel!= -\frac12 \pm \frac{\sqrt5}2$$ So your roots are: $$\{x | p(x) = 0\} = \{0\} \cup \{x|p_1(x) = 0\} = \{0,1\} \cup \{x|p_2(x) = 0\} = \{0,1,-\frac12+\frac{\sqrt5}2, -\frac12 -\frac{\sqrt5}2\}$$

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After factoring out $x$ you obtain a cubic whose highest and lowest coefficients are $\color{#c00}{\pm1}.\,$ If it is reducible then it has a rational root, so by the Rational Root Test, that root can only be $\color{#c00}{\pm1}$. Thus for polynomials of this form you need only evaluate it a two points to complete the factorization.

Generally one can use the Rational Root Test to reduce the factorization of cubics to the testing of a finite number of rational roots, but there will be more than $2$ possible roots to check when the polynomial's high and low coeff's have many factors.

More generally, one can reduce polynomial factorization to the integer factorization of a few values taken by the polynomial, using an old algorithm due to Bernoulli, Schubert, Kronecker, Hausman. However, this is not efficient compared to modern algorithmic methods.

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Where does it say the domain of factorization is $\mathbb Q$? Usually it is $\mathbb R$ and the present polynomial appears to be reducible over $\mathbb R$. – AlexR Jan 30 '14 at 17:51
@AlexR Right, the domain is not specified in the OP. It could be over $\Bbb Q$, over its algebraic closure, or over $\Bbb R$. – Bill Dubuque Jan 30 '14 at 17:57