Solve for $x$ in this equation How do I solve for $x$ algebraically?
$$\dfrac{x^2(x^2-1)}{x+3} = 12$$

Rearranging the equation, we get $$x^2(x^2-1) = 12(x+3)$$
$$x^4 - x^2 - 12x - 36 = 0$$
First we search for integer roots. The integer root must divide $36$. Hence, the possible integer options are $\pm 1,\pm 2, \pm 3$. Checking these $6$ options, give us $x=-2$ and $x=3$. Hence, $$x^4 - x^2 - 12x - 36 = (x+2)(x-3)(x^2+ax+b)$$
Comparing coefficients, we get $a=1$ and $b=6$. Solving the quadratic, gives the other roots as $$x^2 + x + 6 =0 \implies \left(x + \dfrac12 \right)^2 + 6 - \dfrac14 = 0 \implies x = -\dfrac12 \pm i \dfrac{\sqrt{23}}2$$
