polynomials of degree 3 over finite field Is there conditions that the polynomials of degree 3, ($f(x)=ax^3+bx^2+cx+d$) have roots in finite field $F_q$? What is the conditions that the polynomial $2x^3-dx^2+1$ have root on $F_p$? Please help or hints me.
 A: Your question is rather strongly Galois theoretic, but after all, finite fields belong also to Galois theory ... 
Let us recall briefly the general setting for the determination of the Galois group $G=Gal(N/K)$ of the splitting field $N$ of a given irreducible polynomial $f\in K[X]$, where $K$ is a field of characteristic $\neq 2$. Thinking of $G$ as a permutation group of the roots $x_j$ of $f$, an important invariant is the discriminant $D$ of $f$, defined as $D=\Delta^2$, where $\Delta$ is the product of the factors $(x_i - x_j)$ over all the pairs $(i, j)$ s.t. $i<j$.  The origin of the discriminant lies in the explicit expressions for the roots of a quadric, or of a cubic (see Cardano's formulas e.g. in Wikipedia). Obviously $D\in K$, and a permutation $s$ of the roots is even (resp. odd) if $s(\Delta)=\Delta$ (resp. $-\Delta$) (the condition $char(K)\neq 2$ intervenes here), so $K(\Delta)$ is the fixed field of the subgroup  of $G$ consisting of even permutations.
If we specialize  to a cubic $f$, $G$ is a subgroup of $S_3$, hence the Galois group of a separable irreducible cubic is either $A_3$ or $S_3$, and it is $A_3$ iff $\Delta \in K$, i.e. $D$ is a square in $K$. In the general case  $f=ax^3+bx^2+cx+d$, a boring computation using the symmetric functions of the roots gives $D=b^2c^2+18abcd-27a^2d^2 -4ac^3 -4b^3d$. Note that in characteristic $\neq3$, a simple linear change of variables will reduce to the case of a trinomial $f=X^3+bX+c$, in which case the discriminant takes the classical form $D=-4b^3-27c^2$.
Back to your question: a cubic $f$ is irreducible over $K$ iff it has no root in $K$; if moreover $K$ is a finite field, any extension of $K$ of finite degree is Galois, so the splitting field $N$ of an irreducible cubic has degree $3$ over $K$. All in all, the answer is: a separable cubic has a root in a finite field $K$ iff its discriminant $D$ is not a square in $K$. Finally the special cubic that you mention over $\mathbf F_p$ boils down to a non quadratic residue condition if $p$ is odd. The case $p=2$ is obvious  ./.  
