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 $t$
 A: You wrote

I have that $20+t\cdot 5 \cdot 24\equiv 0$ mod $25$ and I want to solve for $t$

So you have $25|20+t\cdot5\cdot24$.  
That means $5|4+t\cdot24$ or $4+t\cdot24\equiv0\pmod5$.  
Can you take it from there?

Addendum:  
The answer above addresses only where OP got stuck.  It does not address other parts of the whole problem.  In fact, as seen in comments and discussion there, $f(4)$ is actually $\equiv0$, not $20,\mod 25,$ in contrast to what the question states, so applying the above would lead to the wrong answer; rather, the approach indicated above should be applied with the correct congruence in order to obtain the correct answer.
A: $x^3+x+57\equiv 0\pmod {125}\Rightarrow x^3+x-68\equiv 0\pmod {125}\Rightarrow (x-4)(x^2+4x+17)\equiv 0\pmod {125}.$ 
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 $\pmod 5.$
