Prove that every irreducible cubic monic polynomial over $\mathbb F_{5}$ has the form $P_{t}(x)=(x-t_{1})(x-t_{2})(x-t_{3})+t_{0}(x-t_{4})(x-t_{5})$? For a parameter $t=(t_{0},t_{1},t_{2},t_{3},t_{4},t_{5},)\in\mathbb F_{5}^{6}$ with $t_{0}\ne 0$ and {$t_{i},i>0$} are ordering of elements in $\mathbb F_{5}$ (t1~t5 is a permutation of [0]~[4] here at least as I think), define a polynomial 
$$P_{t}(x)=(x-t_{1})(x-t_{2})(x-t_{3})+t_{0}(x-t_{4})(x-t_{5}).$$


*

*Show that $P_{t}(x)$ is irreducible in $\mathbb F_{5}[x]$.

*Prove that two parameters $t,t'$ give the same polynomial over $\mathbb F_5$ if and only if $t_{0}=t_{0}'$ and $\{t_{4},t_{5}\}=\{t_{4}',t_{5}'\}$.

*Show that every irreducible cubic monic polynomial over $\mathbb F_{5}$ is obtained in this way.
After trying  $x,x-1,x-2,x-3,x-4$ the first question  can be solved. But I have no idea about where to start with the remaining two. Expanding the factor seems failed for proving two polynomials are equal to each other.

Hints:


*

*A cubic is reducible, only if it has a linear factor. But then it should have a zero in $\mathbb{F}_5$, so it suffices to check that none of $t_1,t_2,t_3,t_4,t_5$ is a zero of $P_t(x)$.

*This part is tricky. I would go about it as follows. Let $t$ and $t'$ be
two vectors of parameters. Consider the difference
$$
Q_{t,t'}(x)=P_t(x)-(x-t'_1)(x-t'_2)(x-t'_3).
$$
It is a quadratic. Show that if $\{t_1,t_2,t_3\}=\{t'_1,t'_2,t'_3\}$ then $Q_{t,t'}$ has two zeros in $\mathbb{F}_5$, but otherwise it has one or none. This allows you to make progress.

*Count them! The irreducible cubics are exactly the minimal polynomials of those elements of the finite field $L=\mathbb{F}_{125}$ that don't belong to the prime field. The number of such elements is $125-5=120$. Each cubic has three zeros in $L$ (it's Galois over the prime field), so there are a total of 40 irreducible cubic polynomials over $\mathbb{F}_5$. How many distinct polynomials $P_t(x)$ are there?

