We've updated our

TEXT

# Factoring a Four Term Polynomial by Grouping

### Learning Outcomes

• Factor a four term polynomial by grouping terms
When we learned to multiply two binomials, we found that the result, before combining like terms, was a four term polynomial, as in this example: $\left(x+4\right)\left(x+2\right)=x^{2}+2x+4x+8$. We can apply what we have learned about factoring out a common monomial  to return a four term polynomial to the product of two binomials. Why would we even want to do this?
Why Should I Care?
Because it is an important step in learning techniques for factoring trinomials, such as the one you get when you simplify the product of the two binomials from above:

$\begin{array}{l}\left(x+4\right)\left(x+2\right)\\=x^{2}+2x+4x+8\\=x^2+6x+8\end{array}$

Additionally, factoring by grouping is a technique that allows us to factor a polynomial whose terms don't all share a GCF. In the following example, we will introduce you to the technique. Remember, one of the main reasons to factor is because it will help solve polynomial equations.

### Example

Factor $a^2+3a+5a+15$

Answer: There isn't a common factor between all four terms, so we will group the terms into pairs that will enable us to find a GCF for them.  For example, we wouldn't want to group $a^2\text{ and }15$ because they don't share a common factor.

$\left(a^2+3a\right)+\left(5a+15\right)$

Find the GCF of the first pair of terms.

$\begin{array}{l}\,\,\,\,a^2=a\cdot{a}\\\,\,\,\,3a=3\cdot{a}\\\text{GCF}=a\end{array}$

Factor the GCF, a, out of the first group.

$\begin{array}{r}\left(a\cdot{a}+a\cdot{3}\right)+\left(5a+15\right)\\a\left(a+3\right)+\left(5a+15\right)\end{array}$

Find the GCF of the second pair of terms.

$\begin{array}{r}5a=5\cdot{a}\\15=5\cdot3\\\text{GCF}=5\,\,\,\,\,\,\,\end{array}$

Factor $5$ out of the second group.

$\begin{array}{l}a\left(a+3\right)+\left(5\cdot{a}+5\cdot3\right)\\a\left(a+3\right)+5\left(a+3\right)\end{array}$

Notice that the two terms have a common factor $\left(a+3\right)$.

$a\left(a+3\right)+5\left(a+3\right)$

Factor out the common factor $\left(a+3\right)$ from the two terms.

$\left(a+3\right)\left(a+5\right)$

Note how the a and 5 become a binomial sum, and the other factor.  This is probably the most confusing part of factoring by grouping.

[latex-display]a^2+3a+5a+15=\left(a+3\right)\left(a+5\right)[/latex-display]

Notice that when you factor a two term polynomial, the result is a monomial times a polynomial. But the factored form of a four-term polynomial is the product of two binomials. As we noted before, this is an important middle step in learning how to factor a three term polynomial. This process is called the grouping technique. Broken down into individual steps, here's how to do it (you can also follow this process in the example below).
• Group the terms of the polynomial into pairs that share a GCF.
• Find the greatest common factor and then use the distributive property to pull out the GCF
• Look for the common binomial between the factored terms
• Factor the common binomial out of the groups, the other factors will make the other binomial
Let’s try factoring a few more four-term polynomials. Note how there is a now a constant in front of the $x^2$ term. We will just consider this another factor when we are finding the GCF.

### Example

Factor $2x^{2}+4x+5x+10$.

Answer: Group terms of the polynomial into pairs.

$\left(2x^{2}+4x\right)+\left(5x+10\right)$

Factor out the like factor, $2x$, from the first group.

$2x\left(x+2\right)+\left(5x+10\right)$

Factor out the like factor, $5$, from the second group.

$2x\left(x+2\right)+5\left(x+2\right)$

Look for common factors between the factored forms of the paired terms. Here, the common factor is $(x+2)$. Factor out the common factor, $\left(x+2\right)$, from both terms.

$\left(x+2\right)\left(2x+5\right)$

The polynomial is now factored.

[latex-display]\left(x+2\right)\left(2x+5\right)[/latex-display]

Another example follows that contains subtraction. Note how we choose a positive GCF from each group of terms, and the negative signs stay.

### Example

Factor $2x^{2}–3x+8x–12$.

$(2x^{2}–3x)+(8x–12)$

Factor the common factor, $x$, out of the first group and the common factor, $4$, out of the second group.

$x\left(2x–3\right)+4\left(2x–3\right)$

Factor out the common factor, $\left(2x–3\right)$, from both terms.

$\left(x+4\right)\left(2x–3\right)$

[latex-display]\left(x+4\right)\left(2x-3\right)[/latex-display]

The video that follows provides another example of factoring by grouping. https://youtu.be/RR5nj7RFSiU In the next example, we will have a GCF that is negative.  It is important to pay attention to what happens to the resulting binomial when the GCF is negative.

### Example

Factor $3x^{2}+3x–2x–2$.

$\left(3x^{2}+3x\right)+\left(-2x-2\right)$

Factor the common factor  $3x$ out of first group.

$3x\left(x+1\right)+\left(-2x-2\right)$

Factor the common factor $−2$ out of the second group. Notice what happens to the signs within the parentheses once $−2$ is factored out.

$3x\left(x+1\right)-2\left(x+1\right)$

Factor out the common factor, $\left(x+1\right)$, from both terms.

$\left(x+1\right)\left(3x-2\right)$

[latex-display]\left(x+1\right)\left(3x-2\right)[/latex-display]

In the following video we present another example of factoring by grouping when one of the GCF is negative. https://youtu.be/0dvGmDGVC5U Sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.

### Example

Factor $7x^{2}–21x+5x–5$.

$\left(7x^{2}–21x\right)+\left(5x–5\right)$

Factor the common factor $7x$ out of the first group.

$7x\left(x-3\right)+\left(5x-5\right)$

Factor the common factor $5$ out of the second group.

$7x\left(x-3\right)+5\left(x-1\right)$

The two groups $7x\left(x–3\right)$ and $5\left(x–1\right)$ do not have any common factors, so this polynomial cannot be factored any further.

$7x\left(x–3\right)+5\left(x–1\right)$

Cannot be factored

In the example above, each pair can be factored, but then there is no common factor between the pairs! In the next section, we will see how factoring by grouping can be used to factor a trinomial.

## Contribute!

Did you have an idea for improving this content? We’d love your input.