<s>[INST] System of equations from roots of polynomial I'm given the equation $3072x^4-2880x^3+840x^2-90x+3=0$ and told that its roots are $\alpha, \alpha r, \alpha r^2, \alpha r^3,$ for some $r\in \mathbb{R}$.
By considering the sum of the roots, the product, etc. I've found that \begin{gather}\alpha(1+r+r^2+r^3)=\frac{15}{16} \\ \alpha^2r(1+r+2r^2+r^3+r^4)=\frac{35}{128} \\ \alpha^3 r^3(1+r+r^2+r^3)=\frac{15}{512} \\ \alpha^4 r^6=\frac{1}{1024}\end{gather}
But this looks like a rather complex system and I can't see any obvious way solve this for $\alpha$ and $r$.
How can this system be solved?
EDIT
I can see that all of the denominators are powers of $2$, but I can't see how that will help me here.
 [/INST] Answer :
$3072x^4-2880x^3+840x^2-90x+3=0$
Factoring by 3
$\Rightarrow $
$1024x^4-960x^3 +280x^2 - 30x+1=0 $
We can see $\frac{1}{2} $ is solution of the equation
$\Rightarrow $
$\frac{1024x^4-960x^3 +280x^2 - 30x+1}{x-\frac{1}{2}} $=0
$\Rightarrow $
$1024x^3 - 448 x^2 + 56 x - 2=0$
We can see $\frac{1}{4}$ is solution of the equation
$\Rightarrow $
$\frac{1024x^3 - 448 x^2 + 56 x - 2}{x-\frac{1}{4}} =0$
$\Rightarrow $
$1024x^2 - 192 x + 8=0$
$\triangle=192^2 - 32(1024)=4096$
$x_1=\frac{192-\sqrt{4096}}{2(1024)}$
And:
$x_2=\frac{192+\sqrt{4096}}{2 (1024) }$
$\Rightarrow $
$x_1=\frac{1}{16}$
$x_2=\frac{1}{8}$
No forget $\frac{1}{2} $also is a solution of the equation
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