Prove that every irreducible cubic monic polynomial over $\mathbb F_{5}$ has the form $P_{t}(x)=(x-t_{1})(x-t_{2})(x-t_{3})+t_{0}(x-t_{4})(x-t_{5})$?

Question

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 $$P_{t}(x)=(x-t_{1})(x-t_{2})(x-t_{3})+t_{0}(x-t_{4})(x-t_{5}).$$

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

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

3. 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.

Answer 1

Hints:

1. 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 $P_t(x)$.
2. 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.
3. 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?