This is an application of the Pigeonhole Principle. The idea is that since there are $n$ possible remainders when an integer is divided by $n$ that at least two of the $n + 1$ integers in the set $\{x_0, x_1, \ldots, x_n\}$ must have the same remainder when divided by $n$. If they have the same remainder when divided by $n$, their difference is divisible by $n$.
The possible remainders when an integer is divided by $n$ are $0, 1, 2, \ldots, n - 1$. If you have $n$ integers, then it is possible for each of them to have a different remainder when divided by $n$. However, if you have $n + 1$ integers, at least two of them must have the same remainder when divided by $n$. Hence, in the set $\{x_0, x_1, \ldots, x_n\}$, there are two numbers $x_i$ and $x_j$, with $i \neq j$, that have the same remainder when they are divided by $n$. Thus, there exist $k, m, r \in \mathbb{N}$, with $0 \leq r \leq n - 1$, such that
\begin{align*}
x_i & = kn + r\\
x_j & = mn + r
\end{align*}
If we take their difference, we obtain
$$x_i - x_j = (kn + r) - (mn + r) = kn - mn = (k - m)n$$
Therefore, $x_i - x_j$ is divisible by $n$.
That AC-method reduces to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $\color{#c00}{=1})$ as follows
$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\!\!\rm\: \color{#c00}{X^2} + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\overbrace{ac,}^{\rm\qquad\ \ \ \ \ {\bf\large\ \ AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \ X = a\:x \\
\end{eqnarray}$$
In your case
$$ {\begin{eqnarray}
f \, &\,=\,& \ \ \, 6 x^2+\ 11\ x\,\ -\ \ 35\\
\Rightarrow\,\ 6f\, &\,=\,&\!\,\ (6x)^2\! +11(6x)-210\\
&\,=\,& \ \ \ \color{#c00}{X^2}+\, 11\ X\,\ -\ 210,\,\ \ X\, =\, 6x\\
&\,=\,& \ \ (X+21)\ (X-\,10)\\
&\,=\,& \ (6x+21)\,(6x-10)\\
\Rightarrow\ \ f\:=\: \color{#0a0}{6^{-1}}\,(6f)\, &\,=\,& \ (2x+\,\ 7)\ (3x\,-5)\\
\end{eqnarray}}$$
In the final step we cancel $\,\color{#0a0}6\,$ by cancelling $\,3\,$ from the first factor, and $\,2\,$ from the second.
If we denote our factoring algorithm by $\,\cal F,\,$ then the above transformation is simply
$$\cal F f\, = a^{-1}\cal F\, a\,f\quad\,$$
Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. The same idea also works for higher degree polynomials, see this answer, which also gives links to closely-related ring-theoretic topics.
Best Answer
We have $$x^2+9x+20$$ We can factorise this to $$(x+4)(x+5)$$ Take $r=x+4$. We now have $$r(r+1)$$For some integer input $x$, and as a result, a integer input $r$, we have an expression for the product of two consecutive integers. One of these must be odd and the other must be even. We know that an even integer multiplied by an odd integer must be even, and so the output of your expression must be even too.