So, since we have a cubic equation with 4 terms, the first thing we should try is factoring by grouping, so:
[tex] 12x^3+2x^2-30x+5 =0 \implies \\
2x^2(6x+1)-5(6x+1)=0 \implies \\
(2x^2-5)(6x+1) = 0 [/tex]
Now that we've factored our equation, we can use ZPP and break it up:
[tex] 2x^2-5 = 0 \implies \\
2x^2=5 \implies\\
x^2=\frac{5}{2} \implies\\
x=\pm\frac{\sqrt{5}}{\sqrt{2}}=\pm\frac{\sqrt{10}}{2} \\
\\
6x+1 =0 \implies \\
6x=-1 \implies\\
x=\frac{-1}{6}
[/tex]
So, our solutions are:
[tex] x\in \{\frac{\sqrt{10}}{2},-\frac{\sqrt{10}}{2},-\frac{1}{6}\} [/tex]
