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*}

### 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*}

**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*}

**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*}

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

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

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

- \begin{align*}3x^2+8x+4\end{align*}
3x2+8x+4

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*}

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

- \begin{align*}6x^2-11x+4\end{align*}
6x2−11x+4

\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*}

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

Factor by grouping. Factor a \begin{align*}3x\end{align*}

\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*}

Yes, this expression can be factored by grouping. First, find \begin{align*}a\cdot c\end{align*}

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*}

\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*} 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.