##
Section3.4Factoring by Grouping

ΒΆThe expressions that are factored in this set of examples are not trinomials - they all have one too many terms. The reason we are exploring this skill is that it is a useful process when factoring trinomials where the leading coefficients isn't \(1\text{.}\)

Let's consider the expression \(10xy+15x-6y^2-9y\text{.}\) The factoring by grouping method begins by considering only the first two terms and factoring from them any and all common factors. In this case we could factor out \(5x\) leaving the residual factor \((2y+3)\text{.}\) We now consider the final two terms. Our goal is to factor out something that also leaves the residual factor \((2y+3)\text{.}\) While it's true that we could factor out \(3y\text{,}\) that would leave behind \((-2y-3)\text{.}\) What we want to factor out is \(-3y\) so that the residual factor is the desired \((2y+3)\text{.}\) Altogether, then, the first step in the process is:

\begin{equation*}
10xy+15x-6xy-9y=\highlightb{5x}\highlight{(2y+3)}\highlightg{-3y}\highlight{(2y+3)}\text{.}
\end{equation*}

The expression now has two terms, with the terms delineated by the subtraction sign. Each of the terms has a factor of \((2y+3)\text{.}\) Just like we can factor (on the right), say, \(z\) from the expression \(\highlightb{5x}\highlight{z}-\highlightg{3y}\highlight{z}\) resulting in \((\highlightb{5x}\highlightg{-3y})\highlight{z}\text{,}\) we can factor \(\highlight{(2y+3)}\) for the expression \(\highlightb{5x}\highlight{(2y+3)}\highlightg{-3y}\highlight{(2y+3)}\) resulting in \((\highlightb{5x}\highlightg{-3y})\highlight{(2y+3)}\text{.}\)

Let's see the process in total:

\begin{align*}
10xy+15x-6y^2-9y\amp=\highlightb{5x}\highlight{(2y+3)}\highlightg{-3y}\highlight{(2y+3)}\\
\amp=(\highlightb{5x}\highlightg{-3y})\highlight{(2y+3)}
\end{align*}

We can use FOIL to check our result.

######
Example3.4.1

Factor \(6x^2y-4x^2+12y-8\text{.}\)

SolutionWe begin by observing that \(2x^2\) can be factored from the first two terms while \(4\) can be factored from the final two terms. Heads up! One of the most common errors made during this process is to forget to put a plus sign in front of the factor of \(4\text{.}\) Written correctly, our factorization process is:

\begin{align*}
6x^2y-4x^2+12y-8\amp=\highlightb{2x^2}\highlight{(3y-2)}\highlightg{+4}\highlight{(3y-2)}\\
\amp=(\highlightb{2x^2}\highlightg{+4})\highlight{(3y-2)}
\end{align*}

######
Example3.4.2

Factor \(7w^2-2w+7w-2\text{.}\)

SolutionWe can see that \(w\) can be factored from the first two terms, but there seems to be nothing that can be factored from the final two terms. When we factor \(w\) from the first two terms, the residual factor is \((7w-2)\) which contains the two terms at the rear of our original expression. The resolution of our dilemma, then, is to factor \(1\) away from the final two terms. We need to remember to put a plus sign in from of that factor of \(1\text{.}\) Going through the complete process we have:

\begin{align*}
7w^2-2w+7w-2\amp=\highlightb{w}\highlight{(7w-2)}\highlightg{+1}\highlight{(7w-2)}\\
\amp=(\highlightb{w}\highlightg{+1})\highlight{(7w-2)}
\end{align*}

Before closing, let's note that \(7w^2-2w+7w-2\) simplifies to \(7w^2+5w-2\text{.}\) This gives us some insight into how the factby grouping process will be useful when factoring trinomials where the leading coefficient isn't \(1\text{.}\)

###
Subsection3.4.1Exercises

###### 1

\(x^2+5x-3x-15\)

Solution\(\begin{aligned}[t]
x^2+5x-3x-15\amp=\highlightb{x}\highlight{(x+5)}\highlightg{-3}\highlight{(x+5)}\\
\amp=(\highlightb{x}\highlightg{-3})\highlight{(x+5)}
\end{aligned}\)

###### 2

\(8x+50+4xy+25y\)

Solution\(\begin{aligned}[t]
8x+50+4xy+25y\amp=\highlightb{2}\highlight{(4x+25)}\highlightg{+y}\highlight{(4x+25)}\\
\amp=(\highlightb{2}\highlightg{+y})\highlight{(4x+25)}
\end{aligned}\)

###### 3

\(4a^2+10a-10a-25\)

Solution\(\begin{aligned}[t]
4a^2+10a-10a-25\amp=\highlightb{2a}\highlight{(2a+5)}\highlightg{-5}\highlight{(2a+5)}\\
\amp=(\highlightb{2a}\highlightg{-5})\highlight{(2a+5)}
\end{aligned}\)

###### 4

\(w^2-6wz-7wz+42z^2\)

Solution\(\begin{aligned}[t]
w^2-6wz-7wz+42z^2\amp=\highlightb{w}\highlight{(w-6z)}\highlightg{-7z}\highlight{(w-6z)}\\
\amp=(\highlightb{w}\highlightg{-7z})\highlight{(w-6z)}
\end{aligned}\)

###### 5

\(xy+5y-2x-10\)

Solution\(\begin{aligned}[t]
xy+5y-2x-10\amp=\highlightb{y}\highlight{(x+5)}\highlightg{-2}\highlight{(x+5)}\\
\amp=(\highlightb{y}\highlightg{-2})\highlight{(x+5)}
\end{aligned}\)

###### 6

\(x^2-11x-9x+99\)

Solution\(\begin{aligned}[t]
x^2-11x-9x+99\amp=\highlightb{x}\highlight{(x-11)}\highlightg{-9}\highlight{(x-11)}\\
\amp=(\highlightb{x}\highlightg{-9})\highlight{(x-11)}
\end{aligned}\)

###### 7

\(3x^2-12x+7x-28\)

Solution\(\begin{aligned}[t]
3x^2-12x+7x-28\amp=\highlightb{3x}\highlight{(x-4)}\highlightg{+7}\highlight{(x-4)}\\
\amp=(\highlightb{3x}\highlightg{+7})\highlight{(x-4)}
\end{aligned}\)

###### 8

\(4y^2-36xy-7xy+63x^2\)

Solution\(\begin{aligned}[t]
4y^2-36xy-7xy+63x^2\amp=\highlightb{4y}\highlight{(y-9x)}\highlightg{-7x}\highlight{(y-9x)}\\
\amp=(\highlightb{4y}\highlightg{-7x})\highlight{(y-9x)}
\end{aligned}\)