Factoring by Grouping

Factoring by grouping is a process where you factor out a common factor from two or more groups of terms. Each grouping of terms also has its own common factor.

Complete the following steps to perform factoring by grouping.

1. Group the terms of the polynomial two by two.

2. Find the GCF of each group.

3. Find a second GCF that is common to the two groups.

After removing the  GCF from each grouping of terms, it is important to ensure that what remains in the groups is identical.

Consider the polynomial |-5xy + x – 20y + 4.|

1. Group the terms of the polynomial two by two
   ||-5xy+x-20y+4=\underbrace{-5xy+x}_{\text{1st group}}\ \ \underbrace{-20y+4}_{\text{2nd group}}||

2. Find the GCF of each group
   1st group:||\begin{align}-5xy+x-20y+4&=\color{green} {\underbrace{-5xy+x}_{\text{1st group}}}\ \ \underbrace{-20y+4}_{\text{2nd group}} \\ &=\color{green}{x(-5y+1)}-20y+4\end{align}||
   2nd group:||\begin{align}-5xy+x-20y+4&=\underbrace{-5xy+x}_{\text{1st group}}\ \ \color{blue}{\underbrace{-20y+4}_{\text{2nd group}}} \\ &=x(-5y+1)\color{blue}{+4(-5y+1)}\end{align}||
   Therefore, the result is |x(-5y+1)+4(-5y+1).|

3. Find a second GCF that is common to the two groups.
   Since |(-5y + 1)| is common to both groups, the factor can be factored out. The result is |(-5y+1)(x+4).|

Consider the polynomial |4x^{3}y + 6x^{2} + 8xy + 12.|

1. Group the terms of the polynomial two by two
   The terms |4x^3y| and |8xy| have two variables in common, thus, it is preferable to group them together since they will likely have more common factors. ||\begin{align}4x^3y+6x^2+8xy+12&=4x^3y+8xy+6x^2+12\\ &=\underbrace{4x^3y+8xy}_{\text{1st group}}\ \ \underbrace{+6x^2+12}_{\text{2nd group }}\end{align}||

2. Find the GCF of each group
   1st group: ||\begin{align}4x^3y+8xy+6x^2+12&=\color{green} {\underbrace{4x^3y+8xy}_{\text{1st group}}}\ \ \underbrace{+6x^2+12}_{\text{2nd group}} \\ &=\color{green}{4xy(x^2+2)}+6x^2+12\end{align}||
   2nd group: ||\begin{align}4x^3y+8xy+6x^2+12&=\underbrace{4x^3y+8xy}_{\text{1st group}}\ \ \color{blue}{\underbrace{+6x^2+12}_{\text{2nd group}}} \\ &=4xy(x^2+2)\color{blue}{+6(x^2+2)}\end{align}|| Therefore, the result is |4xy(x^2+2)+6(x^2+2).|

3. Find a second GCF that is common to the two groups
   Since |(x^2 + 2)| is common to both groups, the factor can be factored out . The result is |(x^2+2)(4xy+6)|. When we factor a polynomial, we generally make sure that it is factored to its simplest form. In the previous example, the second bracket can be factored again using the GCF. ||(x^2+2)(4xy+6)|| Bring out the GCF |2| in the second bracket. ||(x^2+2)(4xy+6)=(x^2+2)\big(\color{red}{2}(\color{red}{2}xy+\color{red}{3})\big)|| The result is |2(x^2+2)(2xy+3).|

It is possible to start by factoring out a GCF and then continue with factoring by grouping using the terms that were obtained. Always check if a GCF can be found before using another factoring method.

Here's how. ||\begin{align}4x^3y+6x^2+8xy+12&=\color{red}{2}\left(\color{red}{2}x^3y+\color{red}{3}x^2 + \color{red}{4}xy + \color{red}{6}\right)\\ &= 2\left(x^2(2xy+3) + 2(2xy+3)\right) \\ &= 2(2xy+3)(x^2+2) \end{align}||

We can expand the factors obtained during the factoring by grouping to check if they are equivalent to the starting polynomial.

Verify the first example ||(-5y+1)(x+4)\overset{?}{=} -5xy-20y+x+4|| Apply the distributive property. ||\begin{align}(-5y+1)(x+4)&\overset{?}{=} -5xy-20y+x+4\\ (-5y\times x)+(-5y\times 4)+(1\times x)+(1\times 4)&\overset{?}{=} -5xy-20y+x+4\\ -5xy-20y+x+4&=-5xy-20y+x+4\end{align}|| The polynomial obtained is equivalent to the starting polynomial, thus confirming the factoring was carried out correctly.