<s>[INST] Solving the equation $2x^4-3x^2-9=0$ How would I go about solving $2x^4-3x^2-9=0$?
I started by taking $x^2$ out, thus getting $x^2(2x^2-3) = 9$ but I don't know if that did anything useful, usually if this equation is equal to $0$ it's easy but this time it's equal to $9$ so I'm lost.
 [/INST] We can add $3x^2-3x^2$ to the left-hand side and factor by grouping.$$\begin{align*}2x^4-6x^2+3x^2-9=0 & \iff 2x^2(x^2-3)+3(x^2-3)=0\\ & \iff (x^2-3)(2x^3+3)=0\end{align*}$$However, if you did not see that sneaky trick, we can always make the substitution $X=x^2$ to get your polynomial into a quadratic.$$2X^2-3X-9=0\iff (X-3)(2X+3)=0\iff (x^2-3)(2x^3+3)=0$$In fact, fourth degree polynomials of the form $ax^4+bx^2+c$ are called biquadratics.
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