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\lhead{\color{blue} \large  Lecture Notes}
\chead{\Large Factoring by Grouping}
\rhead{\small page \thepage}
\lfoot{\small  \copyright  \; \;   Hidegkuti,  2018}
\rfoot{\small Last revised: December 27, 2018}
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\textbf{Grouping} is a factoring technique that can be used in several
situations. \ Grouping consists of strategically pairing (or grouping)
terms, and then factoring out the greatest common factor or GCF three times.
\ We can also say that factoring by grouping is a reversal of FOIL (First,
Outer, Inner, Last). \ We should consider factoring by groupimg when we have
four terms and perhaps also several variables or higher degrees.\vspace{%
0.08in}

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\textbf{Example 1.} \ Completely factor the expression $15ax+6ay-10bx-4by$.

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\textbf{Solution: }\ The first step is grouping the four terms into two
pairs. The goal is to pair terms that have similar terms or coefficients
with common divisors. \ The goal is to have as much of a GCF in a pair as
possible.\vspace{0.08in}

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In this case, we have two options that would both work. \ We could pair $%
15ax $ with $6ay$ because they share the common factor of $3a$. \ We could
also pair $15ax$ with $-10bx$ because then $5x$ is a shared factor. \ The
only pairing that would not work is to pair $15ax$ with $4by$ as these two
terms have no common factor besides $1$.\vspace{0.08in}

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So, we first pair the first two terms and the second two tems. \ As we
stated before, grouping is to factor out the common factor three times. \
First, we factor out the greatest common factor from $15ax+6ay$.%
\begin{equation*}
3a\left( 5x+2y\right) -10bx-4by
\end{equation*}%
This method can only work if, when factoring out the GCF from the second
pair, what is left in the parentheses is identical to $5x+2y$. \ The
greatest common factor of $-10bx-4by$ is $2b$. \ This means that we have two
choices: either factor out $2b$ or $-2b$. \ We have to select the sign that
guarantees that we have $5x+2y$ left in the parentheses. \ Therefore, we
factor out $-2b$ from $-10bx-4by$.%
\begin{equation*}
3a\left( 5x+2y\right) -2b\left( 5x+2y\right)
\end{equation*}%
At this point, $5x+2y$ is the common factor of the two terms and so we can
factor that out.%
\begin{equation*}
3a\underset{\text{GCF}}{\underbrace{\left( 5x+2y\right) }}-2b\underset{\text{%
GCF}}{\underbrace{\left( 5x+2y\right) }}=\left( 5x+2y\right) \left(
3a-2b\right)
\end{equation*}%
So the factored form is \fbox{$\left( 5x+2y\right) \left( 3a-2b\right) $}$.$
\ When we check, we can see why this method is a reversal of FOIL:%
\begin{equation*}
\left( 5x+2y\right) \left( 3a-2b\right) =15ax-10bx+6ay-4by
\end{equation*}

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The next example illustrates a commonly occuring situation, when one or more
of the GCF is $1$. \ Still, grouping will work just fine.\vspace{0.08in}

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\textbf{Example 2.} \ Completely factor the expression $6mx-3x-2m+1$.\vspace{%
0.08in}

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\textbf{Solution: }\ The first step is grouping the four terms into two
pairs. The goal is to pair terms that have similar terms or coefficients
with common divisors. \ The goal is to have as much of a GCF in a pair as
possible. \ The terms here seem to share not much in common, but the first
and third term share $2m$. \ So, we first rearrange the terms:%
\begin{eqnarray*}
6mx-3x-2m+1 &=&6mx-2m-3x+1\text{ \ \ \ \ \ the GCF in the first two terms is 
}2m \\
&=&2m\left( 3x-1\right) -3x+1
\end{eqnarray*}%
The second pair shares no factor besides $1$, but we already see the
similarity between the expression left after factoring out the GCF from the
first pair and the second pair. \ The only quation is, should we factor out $%
1$ or $-1$? \ The goal is to have the same expression left in the
parentheses. \ Therefore, we should factor out $-1$. 
\begin{equation*}
2m\left( 3x-1\right) -3x+1=2m\left( 3x-1\right) -1\left( 3x-1\right)
\end{equation*}%
Now $3x-1$ is the common factor. 
\begin{equation*}
2m\left( 3x-1\right) -1\left( 3x-1\right) =\left( 2m-1\right) \left(
3x-1\right)
\end{equation*}%
So the factored form is \fbox{$\left( 2m-1\right) \left( 3x-1\right) $}$.$ \
When we check, we can see why this method is a reversal of FOIL:%
\begin{equation*}
\left( 2m-1\right) \left( 3x-1\right) =6mx-2m-3x+1
\end{equation*}

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Sometimes we see only one variable, but with higher degrees. \ In this case,
having four terms is still an important indication that grouping would work.%
\vspace{0.08in}

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\textbf{Example 3.} \ Completely factor the expression $2\left(
x^{2}-4\right) \left( 5x+3\right) =\allowbreak 10x^{3}+6x^{2}-40x-24$.%
\vspace{0.08in}

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\textbf{Solution: }\ Like with all factoring, we start with the GCF. \ \ In
this case, all terms are divisible by $2$, so we will factor it out. \ Then
we group the terms by degrees.%
\begin{eqnarray*}
10x^{3}+6x^{2}-40x-24 &=&2\left( 5x^{3}+3x^{2}-20x-12\right) \text{ \ \ \ \
\ \ \ \ \ factor out }x^{2}\text{ from first pair} \\
&=&2\left[ x^{2}\left( 5x+3\right) -20x-12\right] \text{ \ \ \ \ \ \ \ \
factor out }-4\text{ from second pair} \\
&=&2\left[ x^{2}\left( 5x+3\right) -4\left( 5x+3\right) \right] \text{ \ \ \
\ \ factor out \ }\left( 5x+3\right) \\
&=&2\left( x^{2}-4\right) \left( 5x+3\right)
\end{eqnarray*}%
Although we are done with factoring by grouping, the expression $2\left(
x^{2}-4\right) \left( 5x+3\right) $ is not \textit{completely} factored,
because $x^{2}-4$ can be factored by the difference of squares theorem into $%
\left( x+2\right) \left( x-2\right) $. \ So the factored form is \fbox{$%
2\left( x+2\right) \left( x-2\right) \left( 5x+3\right) $}. \ We can check
by multiplying back: \ 
\begin{equation*}
2\left( x+2\right) \left( x-2\right) \left( 5x+3\right) =2\left(
x^{2}-4\right) \left( 5x+3\right) =2\left( 5x^{3}+3x^{2}-20x-12\right)
=10x^{3}+6x^{2}-40x-24
\end{equation*}%
and so our solution is correct.\vspace{0.08in}

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Another important application of grouping is factoring a general quadratic
trinomial, $ax^{2}+bx+c$.\vspace{0.08in}

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\textbf{Example 4.} \ Completely factor $14x^{2}-8x+21x-12$.\vspace{0.08in}

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\textbf{Solution: }\ We group $14x^{2}$ with $21x$ and $-8x$ with $-12$.%
\begin{eqnarray*}
14x^{2}-8x+21x-1 &=&14x^{2}+21x-8x-12\text{ \ \ \ \ \ \ factor out GCF from
first pair} \\
&=&7x\left( 2x+3\right) -8x-12\text{ \ \ \ \ \ \ \ factor out GCF from
second pair} \\
&=&7x\left( 2x+3\right) -4\left( 2x+3\right) \text{ \ \ \ \ factor ou the
GCF \ }2x+3 \\
&=&\left( 2x+3\right) \left( 7x-4\right)
\end{eqnarray*}%
So the factored form is \fbox{$\left( 2x+3\right) \left( 7x-4\right) $}. \
We can check by multiplying back: 
\begin{equation*}
\left( 2x+3\right) \left( 7x-4\right) =14x^{2}-8x+21x-12
\end{equation*}

\pagebreak

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Note that general trinomials usually do not occur with two like terms such
as $-8x+21x$. \ So, how could we factor \newline
$14x^{2}+13x-12$ \ by grouping? \ The trick is to strategically 'take apart'
the linear term $13x$ to two like terms $-8x+21x$. \ But how do we know how
to take apart the linear term? \ There is a systematic way to do that, and
it will be discussed later. \ This factoring technique is called the
AC-method.

\vspace{0.2in}

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Completely factor each of the following.\vspace{0.08in}%
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\begin{enumerate}
\item $5x-6a+2ax-15$

\item $2a-4x^{2}+2ax^{2}-4$

\item $5ax^{3}-5bx^{3}+5ax^{2}y-5bx^{2}y$

\item $3a-3ax-3ay+3axy$

\item $18m-90n-2mx^{2}+10nx^{2}$

\item $p^{2}x^{2}-p^{2}y^{2}-q^{2}x^{2}+q^{2}y^{2}$

\item $2a^{2}by-6a^{2}bt-3a^{2}btx+a^{2}bxy$

\item $2a^{2}b-50b-25bm^{3}+a^{2}bm^{3}$

\item $6x^{5}-15x^{4}+2x-5$

\item $-24ax^{9}+8ax^{7}+6ax^{3}-2ax$

\item $x^{3}+2x^{2}-4x-8$

\item $6x^{2}+16x-3x-8$

\item $2x^{2}+5x-6x-15$

\item $x^{2}-2x-4x+8$

\item $6x^{2}+4x-3x-2$

\item $10x^{2}-25x-4x+10$
\end{enumerate}

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\pagebreak\ \ \ \ 

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\vspace{0.04in}

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'work.jpg';file-properties "XNPEU";}} \ \ \ {\large Practice Problems}%
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\begin{enumerate}
\item $\left( x-3\right) \left( 2a+5\right) $ \ \ \ \ 2. \ $2\left(
a-2\right) \left( x^{2}+1\right) $ \ \ \ \ 3. \ $5x^{2}\left( a-b\right)
\left( x+y\right) $ \ \ \ \ 4. \ $3a\left( x-1\right) \left( y-1\right) $

\item[5.] $2\left( x+3\right) \left( x-3\right) \left( 5n-m\right) $ \ \ \ \
\ \ 6. \ $\left( x+y\right) \left( x-y\right) \left( p+q\right) \left(
p-q\right) $ \ \ \ \ \ 7. \ $a^{2}b\left( x+2\right) \left( y-3t\right) $

\item[8.] $b\left( a+5\right) \left( a-5\right) \left( m^{3}+2\right) $ \ \
\ \ 9. $\ \left( 3x^{4}+1\right) \left( 2x-5\right) $ \ \ \ \ \ 10. \ $%
-2ax\left( 3x^{2}-1\right) \left( 2x^{3}+1\right) \left( 2x^{3}-1\right) $

\item[11.] $\left( x-2\right) \left( x+2\right) ^{2}$ \ \ \ \ \ 12. \ $%
\left( 3x+8\right) \left( 2x-1\right) $ \ \ \ \ \ 13. \ $\left( 2x+5\right)
\left( x-3\right) $ \ \ \ \ \ 14. \ $\left( x-2\right) \left( x-4\right) $

\item[15.] $\left( 3x+2\right) \left( 2x-1\right) $ \ \ \ \ 16. \ $\left(
5x-2\right) \left( 2x-5\right) $\vspace{0.4in}\vspace{4.8in}\vspace{0.4in}
\end{enumerate}

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\end{document}