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I'm having trouble solving the following equality: $$x^5 + 25x^4 + 250x^3 + 1250x^2 + 3124x + 3120 $$ $$=$$ $$ax^5 + (b+c)x^4 +(10c+3d)x^3 + (d(b+c))x^2 + (24a+62d)x + 104e$$

It's supposed to end up in an augmented matrix form, so I can solve it with gaussian elimination and the likes, but it has been forever since I've dealt with this. Any tips on how to begin? I also couldn't think of more relevant search topics, so I apologize.

I recall something to do with, instead of just equating the coefficients of the terms with the same exponents, I should instead factor out the variables I want to find out. So I did that, but couldn't go much further.

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  • $\begingroup$ Yep, but I am supposed to use algorithms like gauss-jordan method to solve this, so I am wondering how would I put this stuff in matrix form such that the solution isn't just immediate. $\endgroup$
    – TestinNowMate
    Oct 4, 2016 at 23:14

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I don't think this'll turn up linear immediately because of the $d(b+c)$ term. If you have to end up with some kind of linear system in matrix form, you could do this:

First off, notice that $b+c = 25$ by equating the $x^4$ coefficients. Then leave the $b+c$ with the $x^4$ untouched, but change $d(b+c)$ to $25d$.
If you now want a matrix-form of this system, I suppose you could plug in $5$ values of $x$ in both sides, and then write that out.

For example taking $x = 1 $ in both sides leads to $7770 = 25 a+b+11 c+90 d+104 e $, so the first row of your matrix would be $\left(25,1,11,90,104\right)$.

$$ \begin{pmatrix} 25 & 1 & 11 & 90 & 104 \\ ? & ? & ? & ? & ? \\ ? & ? & ? & ? & ? \\ ? & ? & ? & ? & ? \\ ? & ? & ? & ? & ? \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \\ e \end{pmatrix} = \begin{pmatrix} 7770 \\ ? \\ ? \\ ? \\ ? \end{pmatrix} $$

Spoiler:

If you want to avoid huge values, try plugging in $x = -6, -5, -4$.

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  • $\begingroup$ Ahhh, exactly what I wanted. Completely forgot plugging values for x! I'll do that then post results. Thanks for now! $\endgroup$
    – TestinNowMate
    Oct 4, 2016 at 23:31
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Substituting $b+c = 25$ from the $x^4$ term for into $d(b+c)$ in the $x^2$ term, we can generate the following matrix.

$\begin{matrix}1& 0& 0& 0& 0&| 1\\ 0& 1& 1&0&0&|25\\ 0&0&10&3&0&|250\\ 0& 0& 0& 25& 0& |1250\\ 24&0&0&62&0&|3124\\ 0&0&0&0&104&|3120 \end{matrix} $

Now put it into Jordan form.

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  • $\begingroup$ I believe I can't do this with this matrix since it is not square. $\endgroup$
    – TestinNowMate
    Oct 4, 2016 at 23:47
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    $\begingroup$ It doesn't need to be square. you can still put it into upper triangular form. When you do your row operations you will produce a 0 vector (hopefully). If you don't (and you do, I checked), then it would indicate that your problem was inconsistent at the set-up. $\endgroup$
    – Doug M
    Oct 4, 2016 at 23:49

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