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Find the equation $$x^5+10x^3+20x-4=0$$

My try:I think this equation maybe take Trigonometric functions

Now I have solution:let

$x=t-\dfrac{2}{t}$,then $$x^5+10x^3+20x-4=(t-\dfrac{2}{t})^5+10(t-\dfrac{2}{t})^3+20(t-\dfrac{2}{t})-4=0$$ so $$t^5-4-\dfrac{32}{t^5}=0$$

My question: why we take $x=t-\dfrac{2}{t}$

and This problem have other solution? Thank you

Glen Wheeler
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  • Can you provide such a $t$? What is a $t$ that makes this a zero of the equation? Here is one value for $x= 0.19620865573750382297$, the other four are imaginary. – Amzoti Oct 30 '13 at 15:21
  • I have edit, Thank you –  Oct 30 '13 at 15:25
  • Because it gives you exact roots as http://www.wolframalpha.com/input/?i=t%5E5-4-32%2Ft%5E5+%3D+0 (I am still not sure how they knew that) – Amzoti Oct 30 '13 at 15:56
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    We take $x=t-\frac 2t$ because it works. I don't see how to guess that it will. The lack of even power terms suggests a substitution like that. You could try $x=t-\frac at$, do the evaluation, and see what value of $a$ makes things cancel nicely. – Ross Millikan Oct 30 '13 at 15:57

1 Answers1

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Here is a thought how you can construct such a polynomial.

Consider the sequence of polynomials $P_n(x)$ with the property $$ P_n\left(t-\frac{2}{t}\right)=t^n+\dfrac{(-2)^n}{t^n}. $$ It is easy to verify that
$P_0=2,\ P_1=x, \ P_2=x^2+4=xP_1+2P_0, \ P_3=x^3+6x=xP_2+2P_1$.

Then we immediately guess a recurrence relation for the polynomials $P_n$: $$P_n=xP_{n-1}+2P_{n-2}.$$ This relation can be easily verified
(i.e. $\left(t-\frac{2}{t}\right)\left(t^{n-1}+\frac{(-2)^{n-1}}{t^{n-1}}\right)+2\left(t^{n-2}+\frac{(-2)^{n-2}}{t^{n-2}}\right)=\left(t^n+\frac{(-2)^n}{t^n}\right)$).

Then we easily compute $P_5(x)=x^5+10x^3+20x$ and therefore
$$P_5\left(t-\frac2t\right)-4=t^5-\frac{32}{t^5}-4.$$ Using the polynomials $P_n$ you can produce many similar examples.

P..
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