Solve for $x$ in this equation How do I solve for $x$ algebraically?
$$\dfrac{x^2(x^2-1)}{x+3} = 12$$
 A: Hint: Rewrite as $x^4-x^2-12x-36=0$. Note that by great good "luck" $x=-2$ and $x=3$ are solutions. (We crossed our fingers and hoped for rational solutions. By the Rational Root Theorem, these have to be of the form $p/q$ where $p$ divides $-36$ and $q$ divides $1$.)  If we had not been lucky, we would be faced with a "general" quartic. Finding solutions to a quartic is in general possible, but very painful.  
A: Note that the equation you get after multiplying out can be rearranged as $$x^4-(x^2+12x+36) = x^4-(x+6)^2 = (x^2+x+6)(x^2-x-6) = 0$$
The second factor gives the roots $x=3$ and $x=-2$ - without guessing. You have to spot the form, though.
A: Multiplying both sides by $x+3$ gives $x^2(x^2-1) = 12 (x+3)$. This can be re-written as
$x^4-x^2-12x-36 = 0$. A few guesses shows that $3$ and $-2$ are solutions, so you can factor these out to get a quadratic.
Explicitly, the equation becomes $(x-3)(x+2)(x^2+x+6) = 0$, hence the solutions are $3$, $-2$, and $\frac{1}{2}(-1 \pm i \sqrt{23}$).
(And, of course, we check that none of these are $-3$.)
A: 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 $$x^2 + x + 6 =0 \implies \left(x + \dfrac12 \right)^2 + 6 - \dfrac14 = 0 \implies x = -\dfrac12 \pm i \dfrac{\sqrt{23}}2$$
