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How would I factor to solve for x?

$x^5 + 20x^2 + 5=0 $?

Do I use synthetic division? Is there a faster/easier way?

Do I have to keep plugging in numbers to see if they equal to zero?

Thanks! I'm not asking for full solutions if you don't want to share (but that would be nice) just opinions on what I should do.

Eleven-Eleven
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Jessica
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    It could also be a quadratic times a cubic. – Will Jagy Oct 21 '13 at 00:42
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    Over the reals, this factors as a linear term and two quadratics -- none of them at all nice. – vadim123 Oct 21 '13 at 00:52
  • @StefanSmith, if you ask Wolfram Alpha kindly to factor it for you, it will give you numerical answers, showing one real root and four complex roots which appear to be conjugate pairs. – dfeuer Oct 21 '13 at 01:05
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    @dfeuer I'm sorry but I can't use Wolfram Alpha during tests. That would be nice though! So do you think you could show me exactly what you did? – Jessica Oct 21 '13 at 01:13
  • @vadim123 how did you get this? – Jessica Oct 21 '13 at 01:14
  • Wolfram was my method. Obviously this isn't a suitable solution for an exam, but it may save you wasting time trying to find "nice" factors where none exist. – vadim123 Oct 21 '13 at 01:16
  • @vadim123 it was a question from my textbook, asking us to sketch the function, therefore I wanted to find the x-intercepts, (there's only 1 real root), but there is indeed a "suitable" solution.. – Jessica Oct 21 '13 at 01:21
  • @vadim123 i just don't know what it is.. – Jessica Oct 21 '13 at 01:23
  • @WillJagy how did you solve that? – Jessica Oct 21 '13 at 01:23
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    To sketch the function you don't need to factor the polynomial; just plug in a few points. – vadim123 Oct 21 '13 at 01:24
  • Note that the function is increasing . Since the leading term is $x^5$, this is the end behavior. Since it is increasing it can only cross the x axis once. – Baby Dragon Oct 21 '13 at 01:27
  • @BabyDragon how would you calculate by hand how to find this "approximate root"? – Jessica Oct 21 '13 at 01:34
  • @StefanSmith, there is surely a way, but I don't know nearly enough about such things. – dfeuer Oct 21 '13 at 02:33
  • @dfeuer : actually, there is an easy analytical tool. Any polynomial with real coefficients can be factored into linear factors and irreducible quadratic factors. The polynomial here is increasing in $x$, goes to $\pm \infty$ as $x$ goes to $\pm \infty$ (respectively), so the polynomial has exactly one real root, and there is exactly one linear factor (using the reals, of course). The remaining fourth-degree polynomial must factor into two quadratics. – Stefan Smith Oct 21 '13 at 16:35
  • @StefanSmith, I guess what I'm missing is why the quartic has to factor into real quadratics. – dfeuer Oct 21 '13 at 16:41
  • @dfeuer : by the Fundemental theorem of algebra, the quartic (with no real roots) has to factor into four linear factors (involving complex numbers), in order for the quartic to have real coefficients, the roots must occur in conjugate pairs, and if you multiply two linear factors corresponding to conjugate pairs, you get a quadratic with real coefficients. – Stefan Smith Oct 22 '13 at 21:32

1 Answers1

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It’s irreducible over the rational numbers, by the Eisenstein Criterion. So in particular, it doesn’t have a rational root. You can get an approximate root by hand in various ways, and the method I would use is the Newton-Raphson method.

Lubin
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  • how would this particular method give me an approximate root by hand for x^5 + 20x^2 + 5=0? – Jessica Oct 21 '13 at 01:33
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    @Jessica, why don't you read the article and try it yourself? It's not hard, just tedious. – dfeuer Oct 21 '13 at 01:41
  • @dfeuer ahaha now I dont want to solve this problem nearly as bad anymore :( I give up. – Jessica Oct 21 '13 at 02:53
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    @Jessica: If you are initially intimidated, you can use the freely downloadable software called Pari/GP to factor polynomials. You can also do it at one of the websites like Wolfram Alpha. But remember that for some it is more interesting to look for the theory behind this stuff than just finding an answer. –  Oct 23 '13 at 04:09