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I have the following polynomial that I want to factor $$ \begin{align*} p(x)= &- 236364091 x^{13}- 28363690920 x^{12}- 1487737229594 x^{11}\\ &- 44880832661940 x^{10} - 860924276925225 x^9- 10941278514219396 x^8 \\ &- 92883153994532540 x^7 - 516122234647942344 x^6 - 1766587850009264989 x^5 \\ &- 3172232270188300464 x^4 - 1263789594225408906 x^3 + 3700632893619477564 x^2 \\ &+ 3120329011246701345 x+5179915266025500 \\ \end{align*} $$ I suspect that $(x+24)$ and $(x+25)$ are factors. I'm interested in the remaining polynomial $q(x)$ such that $p(x)=(x+24)(x+25)q(x)$. I've tryed Mathematica with no success. Can anyone help?

I'm also interested in other software packages more suited to the factorization of this kind of polynomials.

Thanks.


I'm back!!!

Unfortunately there was an error in the above polynomial, so it didnt factor as expected. The correct one follows $$ \begin{align*} p(x)= &-2363640910 x^{13}- 283636909200 x^{12}- 14830715886140 x^{11} \\ &-445215783064800 x^{10}- 8492530754498370 x^9- 107310315789497520 x^8 \\ &- 905953817318009480 x^7- 5008244551318927680 x^6- 17061651312407299570 x^5\\ &- 30494240185050800880 x^4- 12007914756669549180 x^3 + 35636084573032450080 x^2\\ &+ 29908147265568403650 x +103598305320510000 \end{align*} $$ and factors as expected to: $$ \begin{align*} p(x)=-10 (24 + x) (25 + x) q(x) \end{align*} $$ where $$ \begin{align*} q(x)=&236364091 x^{11}+ 16781850461 x^{10}+518942461425 x^9\\ &+ 9024287420055 x^8 + 95697515012142 x^7+ 627280891321794 x^6\\ &+ 2440109049747842 x^5+ 4890576901172110 x^4+ 2461433234591367 x^3\\ &- 5532350693162895 x^2- 4983281122883427 x-17266384220085 \end{align*} $$ Thanks, next time I'll think twice...

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This is an evil, devilish polynomial, yet...can't Mathematica check whether $\,-25\,,\,-24\,$ are roots of your polynomial? If they are then dividing by their factors' product should be relatively easy even for less powerful programms... –  DonAntonio Dec 28 '12 at 13:07
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How did you come across this polynomial? –  Raskolnikov Dec 28 '12 at 13:09
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Factor over what field ? –  Belgi Dec 28 '12 at 13:15
    
The polynimal evaluates to $22423918741962280042500$ at $x=-24$, and to $75358418484740751360000$ at $x=-25$, so neither $x+24$ nor $x+25$ are factors. The $\gcd$ of these evaluations is $2^2.3^5.5^4.7.11.13.103$. Did you by any chance just know about divisibility modulo some number? –  Marc van Leeuwen Dec 28 '12 at 14:08
    
Hell... I've to check everithing again... It should really be $0$ at $x=-24$ and $x=-25$, I'll be back... –  Neves Dec 28 '12 at 14:30
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1 Answer 1

This function has no factorization in $\mathbb{Q}$. I did a manual check with Mathematica.

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