<s>[INST] Question regarding solving polynomial of congruence? In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:
Let $y = 2x^{3} + 7x - 4$, we have:
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$
What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this?  
Thanks, 
 [/INST] $$\begin{aligned}
2x^3+7x-4 \equiv 0 ( mod 5)\\
2x^3+2x-4 \equiv 0 ( mod 5)\\
2(x-1)(x^2+x+2) \equiv 0 ( mod 5) 
\end{aligned}$$
Now you have two solutions $x-1 \equiv 0 (mod5)$ or $x^2+x+2 \equiv 0 (mod5)$.
You can continue and verify if $x \equiv 1 ( mod 5)$ is the only solution.
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