9.10: Factoring by Grouping
Suppose you belong to your school's math club, and you're having a discussion with your fellow members about how to factor the polynomial \begin{align*}5x^2 + 17x + 6\end{align*}
Guidance
It may be possible to factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factoring by grouping. The following example illustrates how this process works.
Example A
Factor \begin{align*}2x+2y+ax+ay\end{align*}
Solution: There isn't a common factor for all four terms in this example. However, there is a factor of 2 that is common to the first two terms and there is a factor of \begin{align*}a\end{align*}
\begin{align*}2x+2y+ax+ay = 2(x+y) +a(x+y)\end{align*}
Now we notice that the binomial \begin{align*}(x+y)\end{align*}
\begin{align*}(x+y)(2+a)\end{align*}
Our polynomial is now factored completely.
We know how to factor quadratic trinomials \begin{align*}(ax^2+bx+c)\end{align*}
 We find the product \begin{align*}ac\end{align*}
ac .  We look for two numbers that multiply to give \begin{align*}ac\end{align*}
ac and add to give \begin{align*}b\end{align*}b .  We rewrite the middle term using the two numbers we just found.
 We factor the expression by grouping.
Let’s apply this method to the following examples.
Example B
Factor \begin{align*}3x^2+8x+4\end{align*}
Solution: Follow the steps outlined above.
\begin{align*}ac=3 \cdot 4=12\end{align*}
The number 12 can be written as a product of two numbers in any of these ways:
\begin{align*}12&=1 \times 12 && and && 1+12=13\\
12 & =2 \times 6 && and && 2+6=8 \qquad \text{This is the correct choice}.\end{align*}
Rewrite the middle term as: \begin{align*}8x=2x+6x\end{align*}
\begin{align*}3x^2+8x+4 = 3x^2+2x+6x+4\end{align*}
Factor an \begin{align*}x\end{align*}
\begin{align*}x(3x+2)+2(3x+2)\end{align*}
Now factor the common binomial \begin{align*}(3x+2)\end{align*}
\begin{align*}(3x+2)(x+2)\end{align*}
Our answer is \begin{align*}(3x+2)(x+2)\end{align*}
In this example, all the coefficients are positive. What happens if the \begin{align*}b\end{align*}
Example C
Factor \begin{align*}6x^211x+4\end{align*}
Solution: \begin{align*}ac=6 \cdot 4 = 24\end{align*}
The number 24 can be written as a product of two numbers in any of these ways.
\begin{align*}24&=1\times 24 && and && 1+24=25\\
24&=(1) \times (24) && and && (1)+(24)=25\\
24&=2 \times 12 && and && 2+12=14\\
24&=(2) \times (12) && and && (2)+(12)=14\\
24&=3 \times 8 && and && 3+8=11\\
24&=(3) \times (8) && and && (3)+(8)=11 \quad \text{This is the correct choice}.\end{align*}
Rewrite the middle term as \begin{align*}11x=3x8x\end{align*}
\begin{align*}6x^211x+4=6x^23x8x+4\end{align*}
Factor by grouping. Factor a \begin{align*}3x\end{align*}
\begin{align*}3x(2x1)4(2x1)\end{align*}
Now factor the common binomial \begin{align*}(2x1)\end{align*}
Our answer is \begin{align*}(2x1)(3x4)\end{align*}
Guided Practice
Factor \begin{align*}10x^243x+28\end{align*}
Solution:
First, find \begin{align*}a\cdot c\end{align*}
\begin{align*} 280=1\cdot 280 \text{ and } 1+280=281\end{align*}
Since 281 is much more negative than 43, you need to have a pair of factors where one is not so negative. Try:
\begin{align*} 280= 7 \cdot 40 \text{ and } 7+40=47\end{align*}
This is close! Since it is still too negative, you need a factor that is less negative than 40, and one that is slightly more negative than 7. Try:
\begin{align*} 280= 8\cdot 35 \text{ and } 8+35=43 \end{align*}
This works! So, 8 and 35 are the factors needed. Rewrite the middle term as \begin{align*}43x=35x8x\end{align*}
\begin{align*}10x^243x+28=10x^235x+8x+28=5x(2x7)4(2x7)=(5x4)(2x7)
\end{align*}
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK12 Basic Algebra: Factor by Grouping and Factoring Completely (13:57)
Factor by grouping.

\begin{align*}6x^29x+10x15\end{align*}
6x2−9x+10x−15  \begin{align*}5x^235x+x7\end{align*}
 \begin{align*}9x^29xx+1\end{align*}
 \begin{align*}4x^2+32x5x40\end{align*}
 \begin{align*}12x^314x^2+42x49\end{align*}
 \begin{align*}4x^2+25x21\end{align*}
 \begin{align*}24b^3+32b^23b4\end{align*}
 \begin{align*}2m^3+3m^2+4m+6\end{align*}
 \begin{align*}6x^2+7x+1\end{align*}
 \begin{align*}4x^2+8x5\end{align*}
 \begin{align*}5a^35a^2+7a7\end{align*}
 \begin{align*}3x^2+16x+21\end{align*}
 \begin{align*}4xy+32x+20y+160\end{align*}
 \begin{align*}10ab+40a+6b+24\end{align*}
 \begin{align*}9mn+12m+3n+4\end{align*}
 \begin{align*}4jk8j^2+5k10j\end{align*}
 \begin{align*}24ab+64a21b56\end{align*}
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factoring by grouping
It is possible to factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factoring by grouping.Factor by Grouping
Factoring by grouping is a method of factoring a polynomial by factoring common monomials from groups of terms.Grouping Symbols
Grouping symbols are parentheses or brackets used to group numbers and operations.Volume
Volume is the amount of space inside the bounds of a threedimensional object.Image Attributions
Here you'll learn how to use grouping in order to factor quadratics when the leading coefficient is not one.