I want to calculate the linear factorisation of the polynomial $f=x^5+5x^4+10x^3+10x^2+5x+70$ in the ring $\mathbb{F}_{101}[x]$. Using Eisenstein $p=5$ and Gauss' lemma, the polynomial is irreducible in $\mathbb{Q}[x]$. However, using symmetric polynomials (I guess), on could factor $f$ completely into linear factors in $\mathbb{F}_{101}[x]$ (normal).
Finding the first root was easy since $f(1)=0$. Hence, by long division we find $$f=(x-1)(x^4+6x^3+16x^2+26x+31)\in\mathbb{F}_{101}[x].$$ We say that $\alpha_i$, with $i=1,\ldots,4$, are the roots of $x^4+6x^3+16x^2+26x+31$ in $\mathbb{F}_{101}[x]$. Now we use symmetric polynomials to find: $$\sum_{1\leqslant i\leqslant4}\alpha_i=\alpha_1+\alpha_2+\alpha_3+\alpha_4=-6,$$ $$\sum_{1\leqslant i<j\leqslant4}\alpha_i\alpha_j=\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4 $$ $$ =\alpha_1(-6-\alpha_1)+\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4=16, $$ $$ \sum_{1\leqslant i<j<k\leqslant4}\alpha_i\alpha_j\alpha_k=\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_2\alpha_4+\alpha_1\alpha_3\alpha_4+\alpha_2\alpha_3\alpha_4 $$ $$ =\alpha_1(\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4)+\alpha_2\alpha_3\alpha_4=-26 $$ and $$ \alpha_1\alpha_2\alpha_3\alpha_4=31. $$ Substituting the second and fourth result into the third gives $$ \alpha_1(16+\alpha_1(6+\alpha_1)+\alpha_2\alpha_3\alpha_4)=-26+31=5. $$ So $\alpha_1$ could be $\pm1,\pm5$ modulo $101$. (I know that $5\cdot87\equiv0$ mod $101$), hence, we could try to solve $$ \alpha_1^2+6\alpha_1+16+\alpha_2\alpha_3\alpha_4=5 $$ but at this point I don't know what I'm doing anymore and I don't know how to proceed. Any help is appreciated!
EDIT: I actually only want to show that it splits into linear factors, which makes it way easier and I know how to do that by using the fact that $\mathbb{F}^*_{101}$ has cyclic order $100$ and contains elements of order $5$. By that and using the morphism $\varphi:\mathbb{Z}[x]\to\mathbb{Z}[x],\;x\mapsto x+1$, we get $f=\varphi(x^5+69)$ and so on...
But what I actually want to know is whether this method will work for finding its roots in $\mathbb{F}_{101}$.
The thing about fifth roots in the deleted question is relevant, because $101$ is prime. Write $y=x+1$ so that $y^5\equiv-69\equiv 32$
Then $y=2$ is a solution and you will find other solutions by solving $z^5=1$, in which case $y=2z$ will be a solution. Once you have found one non-trivial fifth root of $1$, the others are $z^2, z^3, z^4$