I was working on a problem, and I ended up with $$\begin{cases}p+r=a_3\\s+q+pr=a_2\\qr+ps=a_1\\qs=a_0\end{cases}$$ and I was wondering if there is a general algebraic formula to find unknowns $p,q,r,s$ given $a_3,a_2,a_1,a_0$. Or a polynomial in terms of $a_3,a_2,a_1,a_0$ that can find the unknowns.

I've tried substituting $p$ with $a_3-r$ but that didn't get me anywhere. And I've completely burned myself out trying to find a solutions.

So I was wondering if you can help me solve this problem. Maybe you see somethings that I don't see. I don't have access to resources such as Mathematica and I don't think Wolfram Alpha knows a command for this sort of problem.

  • $\begingroup$ Can you tell us something about the $a_i$? (e.g. $a_i=0$ or $1$ would be helpfull) $\endgroup$ – ctst Sep 19 '16 at 16:17
  • $\begingroup$ you can use the Solve[.] command, Mathematica can solve this problem $\endgroup$ – Dr. Sonnhard Graubner Sep 19 '16 at 16:29
  • $\begingroup$ @Dr.SonnhardGraubner: the OP who doesn't "have access to resources such as Mathematica" must really appreciate your input $\ddot{\sim}$. $\endgroup$ – Rob Arthan Sep 19 '16 at 16:31
  • $\begingroup$ but he knows Wolfram Alpha or have i misreaded it? $\endgroup$ – Dr. Sonnhard Graubner Sep 19 '16 at 16:32
  • $\begingroup$ Apologies if you meant Wolfram Alpha can solve it. If so, I will leave my inappropriate comment, but only for the sake of the smiley. $\ddot{\smile}$. $\endgroup$ – Rob Arthan Sep 19 '16 at 16:34

Assume $p$ given. Then the first equation determines $r$. Then the second and third are linear equations in $s,q$, hence can easily be solved for these (assuming $p\ne r$). Finally check if fourth equation holds.

So: $$r=a_3-p$$ $$ s+q=a_2-pr,\ ps+rq=a_1, \implies q=\frac{pa_2-p^2r-a_1}{p-r},\ s=\frac{ra_2-pr^2-a_1}{r-p}$$ hence $$a_0=qs=\frac{(pa_2-p^2r-a_1)(ra_2-pr^2-a_1)}{(p-r)(r-p)=} $$ This leads to the following messy equation in $p$: $$p^6-3a_3p^5+(3a_3^2+2a_2)p^4-(a_3^3+4a_2a_3)p^3+(2a_2a_3^2+a_1a_3+a_2^2-4a_0)p^2+(-a_1a_3^2-a_2^2a_3+4a_0a_3)p+(a_1a_2a_3a_1^2-a_0a_3^2)=0$$

  • $\begingroup$ The only known values are $a_3,a_2,a_1,a_0$ because they are the coefficients of $x^4+a_3x^3+a_2x^2+a_1x+a_0=0$. I'm trying to factor into $(x^2+px+q)(x^2+rx+s)$... $\endgroup$ – Frank Sep 19 '16 at 17:26
  • $\begingroup$ You have a random equal sign in the denominator of $$\frac {(pa_2-p^2r-a_1)(ra_2-pr^2-a_1)}{(p-r)(r-p)}$$ $\endgroup$ – user332252 Sep 20 '16 at 12:35

for the variable $s$ we find $${a_0} {a_1}-{a_0} {a_2}+{a_0} s-{a_1} s+{a_2} s^2-s^3={a_0} {a_3} (s-1)$$

  • $\begingroup$ For some reason, it doesn't work with $a_3=6,a_2=11,a_1=6,a_0=-24024$. I get the cubic $f(s)=s^3-11s^2-120114s+24024=0$ with a bunch of "complex" solutions. Perhaps you copied the original equation wrong? $\endgroup$ – Frank Sep 20 '16 at 3:43
  • $\begingroup$ How did you get to there? I am interested on how you did that. $\endgroup$ – Frank Sep 20 '16 at 23:00

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