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September 2nd, 2014, 02:54 AM   #1
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Factoring Polynomials


is there an "easy" way to find non-rational factors of polynomials?

What I mean is, if I have a polynomial such as:

$\displaystyle x^2+2x-15$'s not too hard to figure out that it factors to:

$\displaystyle (x-3)(x+5)$

...but, if I have a polynomial such as:

$\displaystyle x^2+9x+9$

...then the factors are much harder to find. I've written a piece of software to find them by brute force (a loop inside a loop that stops when the two loop variables equal the two coefficients in the polynomial), but I'm currently running it on the polynomial above and it's not showing any signs of success yet

<edit> it finished while I was typing this and I accidentally pressed enter and cleared the result - now I'm having to run it again - aargh! <edit>

I know the solution for the above polynomial is:

$\displaystyle (x+7.854)(x+1.146)$

...but how do I get there?

Any suggestions will be very much appreciated, not least by my laptop cpu...

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September 2nd, 2014, 09:23 AM   #2
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...I thought I should probably just say that I realise I can find the roots of a quadratic polynomial via $\displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ but I was looking for a factoring way of working with non-rational numbers, if you see what I mean...
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September 2nd, 2014, 09:35 AM   #3
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You could try completing the square, but that is really the same as
$$x = \tfrac{1}{2a}(-b \pm \sqrt{b^2 - 4ac})$$

For what it's worth:
x^2 + 9x + 9 &= (x + \tfrac92)^2 + (9-\tfrac{81}{4}) = 0 \\
(x + \tfrac92)^2 &= \tfrac{45}{4} \\
x + \tfrac92 &= \pm \sqrt{\tfrac{45}{4}} \\
x &= -\tfrac92 \pm \tfrac32 \sqrt{5} \\

Last edited by v8archie; September 2nd, 2014 at 09:37 AM. Reason: Added the radical symbol
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September 2nd, 2014, 10:09 AM   #4
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What degrees of polynomials are you working with? The answer will be very different if you say "2" than if you say "1000".
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