Question

Find all the zeros of the equation

Answer 1

Divide both sides by -3, and replace $x^2$ with $y$. Then

$-3x^4+27x^2+1200=0\iff y^2-9y-400=0$

Factorize the quadratic in $y$ to get

$y^2-9y-400=(y+16)(y-25)=0\implies y=-16,y=25$

which in turn means

$x^2=-16,x^2=25$

But $x^2\ge0$ for all real $x$, so we can ignore the first solution. This leaves us with

$x^2=25\implies x=\pm\sqrt{25}=\pm5$

If we allow for any complex solution, then we can continue with the solution we ignored:

$x^2=-16\implies x=\pm\sqrt{-16}=\pm i\sqrt{16}=\pm4i$