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.
 A: Since you see powers of $2$,  you can use a factorization:
$$3072x^4-2880x^3+840x^2-90x+3=3(1024x^4-960x^3+280x^2-30x+1)=$$
$$=3(1024x^4-64x^3-896x^3+56x^2+224x^2-14x-16x+1)=$$
$$=3(16x-1)(64x^3-56x^2+14x-1)=$$
$$=3(16x-1)(64x^3-8x^2-48x^2+6x+8x-1)=$$
$$=3(16x-1)(8x-1)(8x^2-6x+1)=2(16x-1)(8x-1)(4x-1)(2x-1).$$
A: Going with your equations... Dividing third equation by first, we get $\alpha^2 r^3=\frac{1}{32}$. This implies $r > 0$ because $\alpha$ must be real as well (follows from the first equation). Plugging this into the second equation we get
$$
\frac{35}{128}=\frac{1}{32}\left(\frac{1}{r^2}+\frac{1}{r}+2+r+r^2\right)=\frac{1}{32}\left(\left(r+\frac{1}{r}\right)^2+\left(r+\frac{1}{r}\right)\right).
$$
Letting $u=r+1/r$ gives quadratic equation
$$
u^2+u-\frac{35}{4}=0.
$$
This yields $u=\frac{5}{2}$ as we must have $u>0$. Then solving corresponding quadratic equation given by $\frac{5}{2}=r+1/r$ we see $r \in \{\frac{1}{2},2\}$. From the first equation we get $\alpha$ and so the two solutions are $r=\frac{1}{2}, \alpha=\frac{1}{2}$ and $r=2, \alpha=\frac{1}{16}$.
Clearly both solutions generate the same set of roots $\{\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16}\}$.
A: You can do away with the denominators by considering equation with reciprocal roots $y=1/x$ which is (dividing by $3$) more convenient to handle
$$ y^4 - 30y^3 + 280y^2 - 960y + 1024=0$$
This is a monic polynomial with integer coefficients whose integer roots can only be factors of constant term $1024$ ie, ${\pm1, \pm2, \pm4, \ldots}$. So we have small list of possible roots and good part is that they are still in G.P.
Check that $2$ is a root and so on.
Indeed we have

*

*$2+4+8+16=30$

*$2\cdot4+2\cdot8+2\cdot16+4\cdot8+4\cdot16+8\cdot16=280$

*$2\cdot4\cdot8+2\cdot4\cdot16+2\cdot8\cdot16+4\cdot8\cdot16=960$

*$2\cdot4\cdot8\cdot16=1024$
Therefore, our original equation has roots ${\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16}}$
A: 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
