## Section4.4Factoring by Grouping

The 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.

###### Example4.4.1.

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

Solution

We 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*}
###### Example4.4.2.

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

Solution

We 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 $1\text{.}$

### ExercisesExercises

###### 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}