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 would it be done?

### Factoring by Grouping

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 Process of Factoring by Grouping

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 should factor by grouping using the following steps:

**Step 1:** We find the product \begin{align*}ac\end{align*}.

**Step 2:** We look for two numbers that multiply to give \begin{align*}ac\end{align*} and add to give \begin{align*}b\end{align*}.

**Step 3:** We rewrite the middle term using the two numbers we just found.

**Step 4:** We factor the expression by factoring out the common binomial.

**Factoring Common Binomials**

The last step in factoring by grouping is to factor out a common binomial. For example, consider the following expression:

\begin{align*}x(3x+2)-5(3x+2)\end{align*}

You can see that the term \begin{align*}(3x+2)\end{align*} appears in both terms of the polynomial. This common term can be factored by writing it in front of a set of parentheses. Inside the parentheses, we write all the terms that are left over when we divide them by the common factor.

\begin{align*}(3x+2)(x-5)\end{align*}

#### Let's factor the following expressions by grouping:

- \begin{align*}2x + 2y + ax + ay\end{align*}

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.

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

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*}.

Note that all the coefficients are positive. What happens if the \begin{align*}b\end{align*} is negative?

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

\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*}.

### Examples

#### Example 1

Earlier, you were asked if the polynomial \begin{align*}5x^2 + 17x + 6\end{align*} can be factored by grouping.

Yes, this expression can be factored by grouping. First, find \begin{align*}a\cdot c\end{align*}: \begin{align*}5\cdot6=30\end{align*}. Since \begin{align*}b=17\end{align*}, the factors of 30 need to add up to 17. Note that the factors will all be positive.

The factor pairs of 30 are 1 and 30, 2 and 15, 3 and 10, or 5 and 6.

\begin{align*}30&=1\times 30 && and && 1+30=31\\ 30&=2\times 15 && and && 2+15=17\\ 30&=3 \times 10 && and &&3+10=13\\ 30&=5 \times 6 && and && 5+6=11\end{align*}

2 and 15 multiply to 30 and add to 17 so they are the factors that we need. We want to rewrite the middle term as \begin{align*}17x = 15x +2x\end{align*} and factor by grouping:

\begin{align*}5x^2+17x+6=5x^2+15x + 2x +6=5x(x+3)+2(x+3)=(5x+2)(x+3) \end{align*}

The answer is \begin{align*}5x^2+17x+6 = (5x+2)(x+3)\end{align*}.

#### Example 2

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

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. Or, you could 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*}

### Review

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*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 9.10.