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

### 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

**Step 1:** We find the product

**Step 2:** We look for two numbers that multiply to give

**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:

You can see that the term

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

2x+2y+ax+ay

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

Now we notice that the binomial

Our polynomial is now factored completely.

3x2+8x+4

Follow the steps outlined above.

The number 12 can be written as a product of two numbers in any of these ways:

Rewrite the middle term as:

Factor an

Now factor the common binomial

Our answer is

Note that all the coefficients are positive. What happens if the

6x2−11x+4

The number 24 can be written as a product of two numbers in any of these ways.

Rewrite the middle term as

Factor by grouping. Factor a

Now factor the common binomial

Our answer is

### Examples

#### Example 1

Earlier, you were asked if the polynomial

Yes, this expression can be factored by grouping. First, find

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

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

The answer is

#### Example 2

Factor

First, find

Since -281 is much more negative than -43, you need to have a pair of factors where one is not so negative. Try:

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.