# 9.10: Factoring by Grouping

**Basic**Created by: CK-12

**Practice**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*}. Someone has suggested factoring by grouping. Do you think that this is a possibility? If so, how could it be done? In this Concept, you'll learn how to factor polynomial expressions by grouping when the leading coefficient isn't 1. Your fellow math club members will be impressed when you explain to them the procedure for using grouping to factor an expression like the one given.

### 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*} that is common to the last two terms. Factor 2 from the first two terms and factor \begin{align*}a\end{align*} from the last two terms.

\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*} is common to both terms. We factor the common binomial and get:

\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*} where \begin{align*}a = 1\end{align*} using methods we have previously learned. To factor a quadratic polynomial where \begin{align*}a \neq 1\end{align*}, we follow the following steps.

- We find the product \begin{align*}ac\end{align*}.
- We look for two numbers that multiply to give \begin{align*}ac\end{align*} and add to give \begin{align*}b\end{align*}.
- 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*} *by grouping*.

**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*}, so the problem becomes the following.

\begin{align*}3x^2+8x+4 = 3x^2+2x+6x+4\end{align*}

Factor an \begin{align*}x\end{align*} from the first two terms and 2 from the last two terms.

\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*} is negative?

#### Example C

*Factor* \begin{align*}6x^2-11x+4\end{align*} *by grouping*.

**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=-3x-8x\end{align*}, so the problem becomes:

\begin{align*}6x^2-11x+4=6x^2-3x-8x+4\end{align*}

Factor by grouping. Factor a \begin{align*}3x\end{align*} from the first two terms and factor –4 from the last two terms.

\begin{align*}3x(2x-1)-4(2x-1)\end{align*}

Now factor the common binomial \begin{align*}(2x-1)\end{align*}.

Our answer is \begin{align*}(2x-1)(3x-4)\end{align*}.

### Guided Practice

*Factor \begin{align*}10x^2-43x+28\end{align*}.*

**Solution:**

First, find \begin{align*}a\cdot c\end{align*}: \begin{align*} 10\cdot28=280\end{align*}. Since \begin{align*}b=-43\end{align*}, the factors of 280 need to add up to -43, so consider pairs of negative factors of 280. There will be a lot of pairs of factors, and you could list them all in order until you find the correct pair, as in Examples B and C. Or, use some number sense to help make the search shorter. Start with -1 as a factor:

\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=-35x-8x\end{align*} and factor by grouping:

\begin{align*}10x^2-43x+28=10x^2-35x+-8x+28=5x(2x-7)-4(2x-7)=(5x-4)(2x-7) \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. CK-12 Basic Algebra: Factor by Grouping and Factoring Completely (13:57)

Factor by grouping.

- \begin{align*}6x^2-9x+10x-15\end{align*}
- \begin{align*}5x^2-35x+x-7\end{align*}
- \begin{align*}9x^2-9x-x+1\end{align*}
- \begin{align*}4x^2+32x-5x-40\end{align*}
- \begin{align*}12x^3-14x^2+42x-49\end{align*}
- \begin{align*}4x^2+25x-21\end{align*}
- \begin{align*}24b^3+32b^2-3b-4\end{align*}
- \begin{align*}2m^3+3m^2+4m+6\end{align*}
- \begin{align*}6x^2+7x+1\end{align*}
- \begin{align*}4x^2+8x-5\end{align*}
- \begin{align*}5a^3-5a^2+7a-7\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*}4jk-8j^2+5k-10j\end{align*}
- \begin{align*}24ab+64a-21b-56\end{align*}

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Here you'll learn how to use grouping in order to factor quadratics when the leading coefficient is not one.