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I need to find the solutions of $$\begin{cases}a+j=4\\aj+b+k=-10\\ak+bj+c+l=-54\\al+bk+cj+d+m=9\\am+bl+ck+dj=226\\bm+cl+dk=125\\cm+dl=-301\\dm=-269\end{cases}$$

To continue with completely factoring an octic. I tried Wolfram Alpha, but it just says that it "doesn't know how to interpret your input".

I tried substituting, but gave up shortly due to how ugly the work became.

Frank
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1 Answers1

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I presume you are trying to solve the octic,

$$P(x) :=x^8 + 4x^7 - 10x^6 - 54x^5 + 9x^4 + 226x^3 + 125x^2 - 301x - 269=0\tag1$$

by equating coefficients with,

$$(x^4 + a x^3 + b x^2 + c x + d)(x^4 + j x^3 + k x^2 + l x + m)=0\tag2$$

You have $8$ equations in $8$ unknowns (correcting a typo in your original post). Since your particular octic has a solvable Galois group, then the unknowns can be given in radicals. Your two quartics are,

$$2x^4 + 2(2 - d)x^3 - (9 + 5d)x^2 + (-11 + 7d)x + (7 + 15d)=0$$

where $d=\pm\sqrt{5}$. To recover the octic, the Mathematica command is,

Resultant[2x^4 + 2(2 - d)x^3 - (9 + 5d)x^2 + (-11 + 7d)x + (7 + 15d), d^2 - 5, d]

which will recompose back the original equation, and which WolframAlpha can do.

Tito Piezas III
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  • @Frank: This kind of system is the long road. The short cut, and what you are trying to do, is to factor the octic over a radical extension. Mathematica has a command for this, Factor[Poly, Extension->Sqrt[5]], but the guys at WA are smart enough to limit the functionality of WA. (Otherwise, Mathematica will be rendered redundant.) – Tito Piezas III Sep 20 '16 at 15:05
  • I just realized this is connected to a previous question you asked. – Tito Piezas III Sep 20 '16 at 15:28
  • xD Yes. I tried finding another way to factor... And got this – Frank Sep 20 '16 at 15:44