With the polynomial $x^4 -7x^3 + 3x^2 + 31x + 20$ where $(x+1)$ is a
factor for example.
Do I use the same procedure but start a degree higher, like
$(x+1)(\alpha x^3 + \beta x^2 + \gamma x + \delta)$ and move on down
to figure out the 2nd degree after that?
There are other techniques, namely the polynomial long division as mentioned in a comment by pedja, the Ruffini's rule and for polynomials with integer coefficients the rational root theorem, but you can apply the method of equating coefficients to
$$\begin{eqnarray*}
P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20 =\left( x+1\right) \left( \alpha x^{3}+\beta x^{2}+\gamma x+\delta \right). \end{eqnarray*}\;\; \tag{1}$$
Expanding the RHS and comparing with the LHS, where the coefficient of $x^{4}$ is $1$, you would conclude that $\alpha =1$ and are left with a simpler equality
$$\begin{eqnarray*}
P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}+\beta x^{2}+\gamma x+\delta \right) \\
&=&x^{4}+\left( \beta +1\right) x^{3}+\left( \beta +\gamma \right)
x^{2}+\left( \gamma +\delta \right) x+\delta .
\end{eqnarray*}$$
Equating coefficients you get the simple system of four linear equations
$$
\begin{equation*}
\left\{
\begin{array}{c}
\delta =20 \\
\gamma +\delta =31 \\
\beta +\gamma =3 \\
\beta +1=-7,
\end{array}
\right.
\end{equation*}$$
whose solution is $\delta =20,\beta =-8,\gamma =11$. So
$$\begin{equation*}
x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left(
x^{3}-8x^{2}+11x+20\right).
\end{equation*}\tag{2}$$
Now by direct inspection (or by the rational root theorem, since $1,2,4$ and $5$ are the positive divisors of $20$) you can find that $x=-1$ is a zero of the cubic polynomial $Q(x):=x^{3}-8x^{2}+11x+20$, ie $Q(-1)=0$. Applying a similar method to $Q(x)$ you would find that
$$\begin{equation*}
Q(x)=x^{3}-8x^{2}+11x+20=\left( x+1\right) \left( x-4\right) \left(
x-5\right) .
\end{equation*}$$
Consequently,
$$\begin{equation*}
x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) ^{2}\left( x-4\right) \left(
x-5\right).
\end{equation*}\tag{3}$$
A shorter method is the Ruffini's rule applied to the polynomial division of $P(x)$ by $(x-r)$. This case (with $P(x)=x^{4}-7x^{3}+3x^{2}+31x+20$ and $r=-1,P(r)=0$, ) is shown bellow, where the remainder is $P(r)=s$.
Best Answer
In this specific case, you have, from your first equation: $$ac=2$$ and, since $a,c\in\mathbb{Z}$, and $2>0$ you have the following possible pairs: $$\begin{array}{|c|c|} \hline a & c\\ \hline 2 & 1\\ \hline 1 & 2\\ \hline -1 & -2\\ \hline -2 & -1\\ \hline \end{array}$$ In the same way, since $b,d\in\mathbb{Z}$, we have that: $$\begin{array}{|c|c|} \hline b & d\\ \hline 3 & -1\\ \hline 1 & -3\\ \hline -1 & 3\\ \hline -3 & 1\\ \hline \end{array}$$ Now, due to the second equation: $$ad+bc=-5$$ we can see that the solution is $(a,b,c,d)=(2,1,1,-3)$.
Now, in general, the problem of finding a factoriasation of $$p(x)=p_0+p_1x+p_2x^2,\ p_0,p_1,p_2\in\mathbb{Z}$$ in the form of: $$E(Ax+B)(Cx+D)$$ with integers coefficients, boils down to finding rational roots of the equation $f(x)=0$, since, if $\frac{a}{b}$ and $\frac{c}{d}$ are the two roots of $f(x)=0$, then: $$p(x)=p_2(x-\frac{a}{b})(x-\frac{c}{d})=\frac{p_2}{ad}(ax-b)(cx-d)$$ So, if $x=\frac{a}{b}$ is a rational root of $p(x)=0$ with $(a,b)=1$, we will have: $$p\left(\frac{a}{b}\right)=0\Rightarrow p_0+p_1\frac{a}{b}+p_2\frac{a^2}{b^2}=0\Rightarrow b^2p_0=a(p_1b+p_2a)\Rightarrow b^2p_0|a\overset{(a,b)=1}{\Rightarrow}p_0|a$$ So, our first result is that $p_0$ should be a divisor of $a$, if $\frac{a}{b}$ is a root of $p(x)=0$. Now, we also note that - in case $a\neq0$, since this is a trivial case, due to the fact that $p_0=0$ and $p(x)=x(p_1+p_2x)$: $$p_0+p_1\frac{a}{b}+p_2\frac{a^2}{b^2}=0\Leftrightarrow p_2+p_1\frac{b}{a}+p_0\frac{b^2}{a^2}=0$$ So, we can see that, as previously, $p_2|b$ is needed, as well.
In this way you can find several similar criteria.
Note: The conditions $p_0|a$ and $p_2|b$ are necessary conditions in order for $p$ to have rational roots, so it does not mean that $p$ has rational roots if $p_0|a$ and $p_2|b$, but that if $p$ has rational roots, then $p_0|a$ and $p_2|b$.