Factoring $2t^4+t^2+2$ into prime polynomials in $\Bbb F_3$

Question

I want to factor $2t^4+t^2+2$ into prime polynomials in $\Bbb F_3$.

I know that in $\Bbb F_3$, we have $2=-1$ so $2t^4+t^2+2=-t^4+t^2-1$ but I can't get any further.

Thanks for your help!

Answer 1

$$2t^4+t^2+2 = -(t^4+2t^2+1) = -(t^2+1)^2$$

Answer 2

The polynomial is biquadratic, with discriminant $(-1)^2-4\cdot 2\cdot 2=0$.

That means it's a perfect square, up to a coefficient: just make it monic using that $1=2\cdot2$, so $$ 2t^4+t^2+2=2(4t^4+2t^2+4)=2(t^4+2t^2+1)=2(t^2+1)^2 $$

_{Note: the quadratic formula holds in all fields of characteristic $\ne2$.}