Extracting integral solutions from a quartic equation The equation
\begin{equation*}
y^{4} + 4y^{3} + 10y^{2} + 12y - 27 = 0
\end{equation*}
has two integral roots. Without resorting to the quartic formula, how would one extract the roots from it?

Here's one way we might proceed:
There is an old and simple result:
Let $R$ be a commutative unital ring and let
$p(x) = \displaystyle \sum_0^n p_i x^i \in R[x]; \tag 1$
then $1$ is a root of $p(x)$ if and only if the sum of the coefficients of $p(x)$ is $0$; that is,
$p(1) = 0 \Longleftrightarrow \displaystyle \sum_0^n p_i = 0; \tag 2$
this is an immediate consequence of
$p(1) = \displaystyle \sum_0^n p_i 1^i = \sum_0^n p_i, \tag 3$
as is easy to see.
We can apply this observation to 
$q(y) = y^4 + 4y^3 + 10y^2 + 12y - 27, \tag 4$
and immediately conclude $1$ is a zero of $q(y)$.  Next we synthetically divide $q(y)$ by $y - 1$, and obtain
$q(y) = y^4 + 4y^3 + 10y^2 + 12y - 27 = (y - 1)(y^3 + 5y^2 + 15y + 27); \tag 5$
We Check:
$(y - 1)(y^3 + 5y^2 + 15y + 27) = y^4 + 5y^3 + 15y^2 + 27y -  y^3 - 5y^2 - 15y - 27$
$= y^4+ 4y^3 + 10y^2 + 12y - 27 = q(y); \tag 6$
setting
$s(y) = y^3 + 5y^2 + 15y + 27, \tag 7$
we now invoke the rational root theorem to limit the integer candidates for a zero of this cubic polynomial to the (integer) divisors of the constant term, that is, to $\pm 1$, $\pm 3$, $\pm 9$, and $\pm 27$; we then make a (hopefully) intelligent guess that
$s(-3) = 0 \tag 8$
might hold; and indeed we see that
$s(-3) = -27 + 45 - 45 + 27 = 0; \tag 9$
thus $-3$ is also a root of $q(y)$; now we (again, synthetically) divide $s(y)$ by $y + 3$:
$s(y) = y^3 + 5y^2 + 15y + 27 = (y + 3)(y^2 + 2y + 9), \tag{10}$
and again we may check:
$(y + 3)(y^2 + 2y + 9) = y^3 + 2y^2 + 9y + 3y^2 + 6y + 27$
$= y^3 + 5y^2 + 15y + 27 = s(y); \tag{11}$
it follows that
$q(y) = (y - 1)(y + 3)(y^2 + 2y + 9); \tag{12}$
we have now extracted two integral roots of $q(y)$; the remaining zeroes satisfy the quadratic equation
$y^2 + 2y + 9 = 0; \tag{13}$
the "formula" yields
$y = \dfrac{-2 \pm \sqrt{-32}}{2} = \dfrac{-2 \pm 4 \sqrt{-2}}{2} = -1 \pm 2i\sqrt 2; \tag{14}$
we see the remaining two roots of $q(y)$ form a complex conjugate pair, not integers!  We have thus verified that $q(y)$ has precisely two integral zeroes, $1$ and $-3$.
