We will factor $6y^{2}+y-7$ where $a=6$, $b=1$, and $c=-7$.AC Method: In this method, we first calculate the product of $a$ and $c$ in $ax^2+bx+c$. For this problem, we have $(6) \cdot (-7) = -42$.The number $-42$ can be factored into the product of two numbers in the following ways: $$\begin{array}{lll}1(-42)&-1(42)&2(-21)&-2(21)\newline3(-14)&-3(14)&6(-7)&-6(7)\newline\end{array}$$ Looking at the corresponding sums, we want a sum of $1$, and this sum happens with $(-6)+7=1$ So we write $$\begin{aligned}[t]6y^{2}+y-7&=6y^{2}-6y+7y-7\newline &=6y\mathopen{}\left(y-1\right)+7\mathopen{}\left(y-1\right)\newline &=\left(6y+7\right)\mathopen{}\left(y-1\right)\end{aligned}$$ Generic Rectangles Method: First, we set up generic rectangles by putting the first term and third term in $6y^{2}+y-7$ into the top left and bottom right rectangles: The first term's coefficient, $6$, can be factored into the product of two numbers (where the first factor is positive) in the following ways: $$\begin{array}{llll}1(6)&2(3)&\end{array}$$ The third term, $-7$, can be factored into the product of two numbers in the following ways: $$\begin{array}{llll}1(-7)&-1(7)&\end{array}$$ We put each pair into the corresponding places next to those generic rectangles, and try to match the area of those rectangles with $6y^{2}+y-7$ by guess-and-check. The following generic rectangles show the solution:Dimensions of those generic rectangles in the diagram give us the solution: $$\begin{aligned}[t]6y^{2}+y-7&=\left(6y+7\right)\mathopen{}\left(y-1\right)\end{aligned}$$