Find all the Zeros and their multiplicities of $f(x)=x^5 +4x^4 +4x^3 -x^2-4x +1$ over $\Bbb Z_5$. 
Find all the Zeros and their multiplicities of $f(x)=x^5 +4x^4 +4x^3 -x^2-4x +1$ over $\Bbb Z_5$.

Firstly,I've found the zeros of $f(x)$,just by simply substituting the elements of $\Bbb Z_5=\{0,1,2,3,4\}$ in $f(x)$.I get $f(1)=0$ and $f(3)=0$.So,by factor theorem $(x-1)$ and $(x-3)$ are factors of $f(x)$.
On dividing $f(x)$ by $(x-1)$, I get $x^4 +5x^3 +9x^2+8x +4$ as quotient and $5$ as remainder,which is equal to zero in $\Bbb Z_5$.So,$f(x)=(x-1)(x^4 +5x^3 +9x^2+8x +4)$,thus,multiplicity of $x=1$ is $1$.
On applying the same procedure for $(x-3)$.I get $f(x)=(x-3)(x^4 +7x^3+25x^2 +74x+218)$,thus,multiplicity of $x=3$ is $1$.
But this is Incorrect.I don't know where I am committing mistake.
Is my approach correct?
If there is any general method for finding the quotient and remainder,when some arbitrary polynomial $T(x)$ is divided by another arbitrary polynomial $g(x)$ in some finite field,Please tell me.
Any suggestions are heartily welcome!!
Thank you!

Start by rewriting $\bar{f}(x)=x^5-x^4-x^3-x^2+x+1$. Divide by $x-1$ to get
$$\bar{f}(x)=(x-1)(x^4-x^2-2x-1)$$
Now divide $x^4-x^2-2x-1$ by $x-3$ to get
$$\bar{f}(x)=(x-1)(x-3)(x^3+3x^2+3x+2)$$
If we consider now $g(x)=x^3+3x^2+3x+2$ we notice that $g(3)=0$. So $3$ is a double root of $f$. Let's divide $g$ by $x-3$ we then get
$$\bar{f}(x)=(x-1)(x-3)^2(x^2+x+1)$$
And we have now completely factored $f$ over $\Bbb{Z}_5$.
We could have as well computed $f'(x)=5x^4-4x^3-3x^2-2x+1$ whose reduction $\pmod 5$ is $x^3+2x^2-2x+1$ and notice that $f'(1)=2$ while $f'(3)=0$ to conclude that $1$ is a simple root and $3$ is a double root.
