8th Grade Algebra Problem I Can't Solve I have no idea how to solve this. Could anyone give a hand? Thanks!
$$2x^{5a} + 6x^{3a} - x^{2a} - 3 = 0$$
PS: I'm guessing this doesn't make sense unless it's also "solve for x" and the equation is = 0, but I'm not certain of that.
PPS: My intuition was to add 3 to both sides, then multiply both sides by x, then divide by 3, but I didn't know where to go from there (or even if that was right).

Edit: To be clear, I'm not the 8th grader. I'm simply unable to help her, and would appreciate assistance. You can view my web dev profile at stackoverflow if you need proof: https://stackoverflow.com/users/174621/matrym
 A: HINT $\ $ It is $\rm\ \ 2\ x^{2a}\ (x^{3a}+3) - (x^{3a} + 3)\:.\ $ Pull out the obvious factor.
Generally one can apply ideas like the Rational Root Test to help find such binomial factors.
A: I am assuming the equation you have written is $2x^{5a} + 6x^{3a} - x^{2a} - 3 = 0$ and you want to solve it.
Let $y = x^a$.
The equation becomes $2y^5 + 6y^3 - y^2 - 3 = 0$.
$$2y^5 + 6y^3 - y^2 - 3 = 2y^3 (y^2 + 3) - (y^2 + 3) = 0$$
$$(2y^3 - 1)(y^2 + 3) = 0$$
Solve for $y$. 
If you are interested only in real solutions, then $y = \frac{1}{\sqrt[3]{2}}$.
Hence, $$x = \frac{1}{\sqrt[3a]{2}}$$
If you are interested in complex roots as well, then you will get $5$ solutions for $y$, namely, $$\frac{1}{\sqrt[3]{2}}, \frac{\omega}{\sqrt[3]{2}}, \frac{\omega^2}{\sqrt[3]{2}}, i\sqrt{3}, -i\sqrt{3}$$
where $\omega$ is the complex cube root of unity and $i^2 = -1$.
Call them $y_1,y_2,y_3,y_4$ and $y_5$.
Then $$x = \zeta \sqrt[a]{y_i}$$ where $\zeta$ is one of the $a$ roots of $a^{th}$ roots of unity and hence you will end up with $5a$ roots.
