<s>[INST] Solving For $X$ In A 4th Degree Polynomial The examples they use on my book are basically from basic arithmetic like 1, 2 and 3 to calculus and calculating derivatives, which is really annoying because I can't build the fundamental skill required to do harder questions. Anyhow, I have to solve an non-factorable inequality. Normally, I would use the quadratic formula but it's not quadratic, nor cubic.
The question is $x^4  + 2x^3  - 4x^2  - 6x \leq -3$. And I have to get an $x$ = something. I am actually really lost here because I can't even apply factor theorem in this case since no values work out.
I think this is a problem of my fundamentals though, please tell me how to solve things like these when they cannot be factored. Or we can just have a one on one discussion...
Any help is appreciated, thank you! 
 [/INST] After manipulating a little bit and rearranging terms when the whole thing is $\leq 0$:
$$x^4  + 2x^3  - 4x^2  - 6x + 3 \\
x^4+2x^3-6x-4x^2+3 \\
x^4 + 2x(x^2-3)-4x^2+3 \\
x^4 + 2x(x^2-3)-3x^2-x^2+3 \\
x^4 + 2x(x^2-3)-3x^2-(x^2-3) \\
x^4 -3x^2 + 2x(x^2-3)-(x^2-3) \\
x^4 -3x^2 + (x^2-3)(2x-1) \\
x^2(x^2-3) + (x^2-3)(2x-1) \\
(x^2-3)(x^2+2x-1).$$
So your problem is now $(x^2-3)(x^2+2x-1) \leq 0$. So it appears that $x=\pm \sqrt{3}$ will bring about equality. If you use the quadratic formula, you will also find that $x=\pm \sqrt{2}-1$ are 2 more solutions. That means we've found all 4 roots(!). 
After testing sample values within each of the intervals delimited by the 4 roots, you can show that the total valid domain is $x \in [-1-\sqrt{2}, -\sqrt{3}] \cup [\sqrt{2}-1,\sqrt{3}]$.
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