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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:

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I imagine the instructions were something besides "solve." Were they something like "simply this expression." Did they give you a value for a? – Michael McGowan Jan 20 '11 at 2:22
Without context (for example, what do the directions say?) it is very hard to interpret what question was intended. – Isaac Jan 20 '11 at 2:24
@Michael: I've relayed all of the information that I have. Is it insufficient? I interpreted the question to be "solve for x"... – Matrym Jan 20 '11 at 2:24
@Matrym: The information you gave is insufficient. For e.g. is it $2x^{5a} + \dots$ or is it $2ax^5 + \dots$? Also I suppose the question is about finding the roots (setting =0 and finding the values of x), but that is not clear. – Aryabhata Jan 20 '11 at 2:27
@Matrym: One cannot "solve" an expression; one usually solves an equation, and an equation requires an "equal" sign somewhere. If the instructions were to "factor", it was unclear; factor into what? But the most likely meaning is they want it to be factored into the terms that are linear in $x^a$. – Arturo Magidin Jan 20 '11 at 2:38

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.

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This looks very promising, thank you! – Matrym Jan 20 '11 at 2:36
Wish I had learned that in 8th grade! – The Chaz 2.0 Mar 14 '11 at 16:00
@The Chaz: At my school, back when I was in the 8th grade (things have changed there since then), even beginning algebra topics (e.g. solve $2x - 1 = 5$ for $x$ or express the area of a rectangle in terms of its length if its width is $3$ more than its length) were not covered in any of the 8th grade math classes. – Dave L. Renfro Nov 7 '11 at 12:42
@DaveL.Renfro - yikes! My father has told me similar stories... – The Chaz 2.0 Nov 7 '11 at 12:51
Nice but no chance at all this is eighth grade solution (the complex solutions part) – Quality Jan 2 at 9:53

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.

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