Strategy for finding integral roots for polynomials with large coefficients I'm trying to find the integer roots for $f(x) = x^5 + 47x^4 + 423x^3 + 140x^2 + 1213x - 420 = 0$. The techniques I'm expected to have at my disposal are:


*

*For a polynomial with integer coefficients $p(x) = a_nx^n + a_{n-1}x^{n-1}+ \dots + a_1x + a_0$, if $p(x)$ has a rational root $\frac{s}{t}$, then $s|a_0$ and $t|a_n$.

*"Einstein's Irreducibility Criterion": If there exists a prime $p$ such that $a_{n-1}\equiv_pa_{n-2}\equiv_p\dots\equiv_pa_0\equiv_p0$, $a_n\not\equiv_p0$ and $a_0\not\equiv_{p^2}0$ then $p(x)$ is irreducible over the rationals.


The book gives the answers $-12$ and $-35$. Using the first strategy, I could have solved for every divisor of $-420$ to find one solution $c$, then divide $(x-c)$ into $f(x)$ and perform the same process with the quotient.  But is there a less time-consuming way to find these roots? I feel like I'm supposed to simplify $f(x)$ by substituting $x$ with something but I wouldn't know what that is.

Try
\begin{align}
  x^5 + 47x^4 + 423x^3 + 140x^2 + 1213x - 420
  & \equiv (x^2+ax+b)(x^3+cx^2+dx+e) \\
  & \equiv x^5+(a+c)x^4+(ac+b+d)x^3 \\
  & \quad +(ad+bc+e)x^2+(ae+bd)x+be
\end{align}
with different combinations of $b$ and $e$
