Solving the congruence $2x^3 - 3x^2 + 1 \equiv 0 \pmod{49}$ I'm trying to solve the congruence $f(x) = 2x^3 - 3x^2 + 1 \equiv 0 \pmod{49}$.
I wrote $f(x)=(x-1)^2(2x+1)$, and I found that it has eight solutions. One solution is $24$, and the others are of the form $7k+1$, where $k = 0, 1, \ldots, 6$.
Are these the only ones?

Yes they are the only solutions.
Your congruence is equivalent to finding $49$ divides $(x-1)^2(2x+1)$ and this implies that:


*

*Either $7$ divides $x-1$ hence $x=1+7k$ which gives you $7$ solutions.

*Or $7$ does not divide $x-1$ and in this case $49$ divides $2x+1$ hence $x\equiv 24 \mod 49$

