Solve:
$3072x^4-2880x^3+840x^2-90x+3=0$ has roots $\alpha$, $r$$\alpha$, $r^2$$\alpha$ and $r^3$$\alpha$
(7 Mark Exam Question)
How would I go about solving this equation using Viete's Formula's. After finding $\sum$$\alpha$, $\sum$$\beta$, $\sum$$\gamma$ and $\sum$$\delta$, I'm not really sure what to do next.
Using Viete's formula's would be preferable as this is what will be on my exam for solving quartics, but if there is a better way of doing it in an exam that would be better.
For class problems, quartics always have rational roots. After you factor out $3$ the leading term becomes $1024x^4,$ the constant $1$, so the possible roots are of the form $\pm \frac 1{2^n}$ with $0 \le n \le 10$. Clearly then $r=\pm \frac 12, \pm \frac 14,$ or $\pm \frac 18$. That is a number to try, but you can get there. In fact, the roots are $\frac 12, \frac 14, \frac 18, \frac1{16}$
To use Vieta, because of the form of the roots you can choose whichever two relations you want. I would use the constant term, which gives $r^6\alpha^4=\frac 1{1024}$ and the $x^3$ term which gives $\alpha \frac {1-r^4}{1-r}=\frac {2880}{3072}=\frac {15}{16}$. Again the solution $r=\alpha=\frac 12$ jumps out because of the $1024=2^{10}$ but grinding through the algebra would be a mess.