Solve $x^3+x+57\equiv 0$ mod $125$ 
Solve $x^3+x+57\equiv 0$ mod $125$

So first I try to solve $x^3+x+2\equiv 0$ mod $5$
I have $f^\prime(x)=3x^2+1$
I found $f(4)$ to be the only solution.
And since $f^\prime(4)\equiv 4$ there is a unique solution mod $25$
So I then need to solve $f(4+t5)=f(4)+t\cdot 5\cdot f^\prime(4)\equiv 0$ mod $25$
$f(4)\equiv 20$ mod $25$ and $f^\prime(4)\equiv 24$ mod $25$
I'm not sure where to go from here,
I have that $20+t\cdot 5 \cdot 24\equiv 0$ mod $25$ and I want to solve for $t$

$x^3+x+57\equiv 0\pmod {125}\Rightarrow x^3+x-68\equiv 0\pmod {125}\Rightarrow (x-4)(x^2+4x+17)\equiv 0\pmod {125}.$ 
How did I get $x-4?$ I simply used the fact that $x-4$ is a factor of the polynomial $x^3+x-68.$ 
Since the quadratic $x^2+4x+17$ has no solutions $\pmod {125},$ the only solution is $x=4\pmod {125}.$ You didn't need to start with $\pmod 5.$
