Suppose that $$x^5+ax^4+bx^3+cx^2+dx+e=0$$ with coef.. all integers, decompose as follows $$(x^2+Ax+B)(x^3+Cx^2+Dx+E)=0$$ which leads to the following system $$ \begin{cases} C+A = a\\ D+AC+B = b\\ E+AD+BC=c\\ AE+BD=d\\ BE=e \end{cases} $$ $$ \begin{cases} C =A- a\\ D+A(A-a)+B = b\\ e/B+AD+B(A-a)=c\\ A(e/B)+BD=d\\ E=e/B \end{cases} $$ $$ \begin{cases} C =A- a\\ A^2-aA+B+D - b=0\\ (A-a)B^2+(AD-c)B+e=0\\ DB^2-dB+Ae=0\\ E=e/B \end{cases} $$ $$A_{1,2}=\frac{a\pm\sqrt{a^2-4(B+D-b)}}{2}$$ $$B_{1,2}=\frac{(AD-c)\pm\sqrt{(AD-c)^2-4(A-a)e}}{2(A-a)}$$ $$B_{1,2}=\frac{d\pm\sqrt{d^2-4ADe}}{2D}$$
How can you go on from there? Is it worth the attempt?
Maybe you have heard that there is no method to solve algebraic equations of degree higher than 4. If your method would work it would allow to solvve equations of 5the degree. you can find one zero point x0 by newtons method for example, and than divide by(x-x0) also with integer coefficients it is often easy to find integer solutions, if the exist.