What if you had a polynomial expression like \begin{align*}3x^2 - 6x + 2x - 4\end{align*} in which some of the terms shared a common factor but not all of them? How could you factor this expression? After completing this Concept, you'll be able to factor polynomials like this one by grouping.

Fisch Video: Factoring Polynomials by Grouping

### Watch This

CK-12 Foundation: 0913S Factoring By Grouping

### Guidance

Sometimes, we can factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called **factor by grouping.**

The next example illustrates how this process works.

#### Example A

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

**Solution**

There is no factor common to all the terms. However, the first two terms have a common factor of 2 and the last two terms have a common factor of \begin{align*}a\end{align*}. 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*}

#### Example B

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

**Solution**

We factor 3x from the first two terms and factor 4 from the last two terms:

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

Now factor \begin{align*}(x+2)\end{align*} from both terms: \begin{align*}(x+2)(3x+4)\end{align*}.

Now the polynomial is factored completely.

### Practice

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*}2a^2-6ab+3ab-9b^2\end{align*}
- \begin{align*}5x^2+15x-2xy-6y\end{align*}

### Additional Practice Opportunities (including self-check)

- Braingenie: Factoring Polynomials by Grouping Terms