Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
easy to see that $x=0$ and $x=1$ are the roots of the given equation. just factor it out. –  Santosh Linkha Jan 30 at 17:34
1  
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 at 17:34
1  
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 at 17:35
1  
The Rational Root Theorem is often useful in school exercises. –  André Nicolas Jan 30 at 17:36
    
@Santosh There are two more roots anyway $\frac12(-1\pm\sqrt5)$ –  Charlie Jan 30 at 17:39
show 1 more comment

2 Answers 2

up vote 2 down vote accepted

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: $$\{x | p(x) = 0\} = \{0\} \cup \{x|p_1(x) = 0\} = \{0,1\} \cup \{x|p_2(x) = 0\} = \{0,1,-\frac12+\frac{\sqrt5}2, -\frac12 -\frac{\sqrt5}2\}$$

share|improve this answer
add comment

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.

share|improve this answer
    
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 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 at 17:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.