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I have an equation $f(x)=x^4+4x^3+2x^22-x+6$. In the past I was taught to factor it by getting the zeros by getting $p/q$, and start guessing zeros, and plugging them into the function. Once I got one or two, I would try to divide the function by them to get the rest.

It would seem to me that there has to be a much easier way of doing this. Some kind of trick. If the equation was only something like $f(x)=x^2+5x-6$, then it would be easy. just find the number that multiplies together to equal $-6$, and adds up to equal $5$. The answer would be $(x+6)(x-1)$. Is there some trick like this for functions with a degree higher than $2$?

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Tricks abound for factoring polynomials of various specific types. However, in general, factoring is quite difficult. The computer algebra algorithms (implemented in Maple/Mathematica etc.) are very complicated and impossible to implement "by hand". If you restrict your attention to quartic polynomials (degree 4), life is still difficult. I can say this with certainty since the quartic equation is quite complicated (although do-able by hand). –  Bill Cook Dec 12 '11 at 23:01
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Nothing really faster here. We are "lucky" that there are the roots $-2$ and $-3$. So divide by $x^2+5x+6$. We get the factorization $(x+2)(x+3)(x^2-x+1)$. Can't go further in the reals, but $x^2-x+1$ factors as $(x-\alpha)(x-\beta)$ where $\alpha=(1+\sqrt{-3})/2\;$, $\beta=(1-\sqrt{-3})/2$. –  André Nicolas Dec 12 '11 at 23:07
    
Actually, I'd change the "impossible" in @Bill's comment to "unwieldy"; otherwise, I quite agree with his assessment. –  J. M. Dec 13 '11 at 0:20

2 Answers 2

up vote 4 down vote accepted

The corresponding "trick" is to just find the four numbers $a,b,c,d$ such that $a+b+c+d=-4$, $ab+ac+ad+bc+bd+cd=2$, $abc+abd+acd+bcd=1$, and $abcd=6$.

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why -4 and +1 instead of +4 and -1? –  Ephraim Dec 12 '11 at 23:01
    
But the OP asked for a method without guessing, which the above is not. –  Bill Dubuque Dec 12 '11 at 23:14
    
Because I was thinking of the roots instead of the factors, that is, I was thinking of $(x-a)(x-b)(x-c)(x-d)$ instead of $(x+a)(x+b)(x+c)(x+d)$. –  Gerry Myerson Dec 12 '11 at 23:15
    
@Bill, it's my way of trying to get OP to see that you can't always get what you want. –  Gerry Myerson Dec 12 '11 at 23:17

It rather depends on what you mean by guessing zeros.

The rational root theorem provides a systematic algorithm for finding rational roots of the form $\frac{p}{q}$: $p$ divides the coefficient of the constant (so is $\pm 1, \pm 2, \pm 3 \text{ or } \pm 6$ in your example), while $q$ divides the coefficient of the highest power (so is $\pm 1$ in your example).

You then test all of the possible pairs (or decide just to stick to positive $q$ to save half the effort) and in your example find that the only rational roots are $-\frac{2}{1}$ and $-\frac{3}{1}$. Then factor these out and check that none of them are repeated zeros.

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I like how a math.SE member put it (paraphrasing): "it sounds dumb, 'try this list of candidates', but at least it's not 'try this infinite list'." –  J. M. Dec 13 '11 at 0:16

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