Learning Outcomes
- Factor a four term polynomial by grouping terms
Key words
- Prime polynomial: a polynomial that does not factor; it is divisible by only 1 and itself.
- Factoring by grouping: a technique to factor polynomials with four or more terms
Factoring by Grouping
When we multiply two binomials, the result, before combining like terms, is a four term polynomial. For example: [latex]\left(x+4\right)\left(x+2\right)=x^{2}+2x+4x+8[/latex].
We can apply what we have learned about factoring out a common monomial to help us factor a four-term polynomial into the product of two binomials.
Why would we even want to do this? Because it is an important step in learning techniques for factoring trinomials, such as the one we get when we simplify the product of the two binomials from above:
[latex]\begin{array}{l}\left(x+4\right)\left(x+2\right)\\=x^{2}+2x+4x+8\\=x^2+6x+8\end{array}[/latex]
Additionally, factoring by grouping is a technique that allows us to factor a polynomial whose terms don’t all share a GCF. We will introduce the technique in the following example. Remember, one of the main reasons to factor is because it will help us solve polynomial equations.
Example
Factor [latex]a^2+3a+5a+15[/latex]
Solution
The only common factor between all four terms is [latex]1[/latex], so we will group the terms into pairs and find a GCF for each pair.
[latex]\left(a^2+3a\right)+\left(5a+15\right)[/latex]
The GCF for the first pair is [latex]a[/latex] and the GCF for the second pair is [latex]5[/latex].
Factor each pair using the GCF:
[latex](a^2+3a)+(5a+15)\\=a(a+3)+5(a+3)[/latex]
Notice that the two terms now have a common factor of [latex]\left(a+3\right)[/latex].
[latex]\color{red}{a}\color{blue}{\left(a+3\right)}\color{green}{+5}\color{blue}{\left(a+3\right)}[/latex]
Factor out the common factor [latex]\left(a+3\right)[/latex] from the two terms.
[latex]\color{blue}{\left(a+3\right)}\left(\color{red}{a}\color{green}{+5}\right)[/latex]
Note how the [latex]a[/latex] and [latex]5[/latex] become a binomial sum, and the other factor.
Answer
[latex]a^2+3a+5a+15=\left(a+3\right)\left(a+5\right)[/latex]
This process is called factoring by grouping. Broken down into individual steps, here’s how to do it (you can also follow this process in the example below).
FACTOR BY GROUPING
- Group the terms of the polynomial into pairs that share a GCF.
- Find the greatest common factor of each pair and use the distributive property to factor each pair
- Look for a common binomial between the factored terms
- Factor the common binomial out of the groups, the other factors will make the other binomial
Example
Factor [latex]2x^{2}+4x+5x+10[/latex].
Solution
Group terms of the polynomial into pairs:
[latex]\left(2x^{2}+4x\right)+\left(5x+10\right)[/latex]
Factor out the like factor, [latex]2x[/latex], from the first group, and [latex]5[/latex] from the second group.
[latex]2x\color{blue}{\left(x+2\right)}+5\color{blue}{\left(x+2\right)}[/latex]
Look for common factors between the factored forms of the paired terms. Here, the common factor is [latex](x+2)[/latex].
Factor out the common factor, [latex]\left(x+2\right)[/latex], from both terms.
[latex]\color{blue}{\left(x+2\right)}\left(2x+5\right)[/latex]
The polynomial is now factored.
Answer
[latex]\left(x+2\right)\left(2x+5\right)[/latex]
Try It
Factor [latex]3x^{2}+6x+7x+14[/latex].
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 [latex]2x^{2}–3x+8x–12[/latex].
Solution
Group terms into pairs:
[latex](2x^{2}–3x)+(8x–12)[/latex]
Factor the common factor, [latex]x[/latex], out of the first group and the common factor, [latex]4[/latex], out of the second group.
[latex]x\left(2x–3\right)+4\left(2x–3\right)[/latex]
Factor out the common factor, [latex]\left(2x–3\right)[/latex], from both terms.
[latex]\left(x+4\right)\left(2x–3\right)[/latex]
Answer
[latex]\left(x+4\right)\left(2x-3\right)[/latex]
Try It
Factor [latex]2x^{2}–3x+10x–15[/latex].
The video that follows provides another example of factoring by grouping.
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 [latex]3x^{2}+3x–2x–2[/latex].
Solution
Group terms into pairs:
[latex]\left(3x^{2}+3x\right)+\left(-2x-2\right)[/latex]
Factor the common factor [latex]3x[/latex] out of first group, and [latex]−2[/latex] out of the second group. Notice what happens to the signs within the parentheses once [latex]−2[/latex] is factored out.
[latex]3x\left(x+1\right)-2\left(x+1\right)[/latex]
Factor out the common factor, [latex]\left(x+1\right)[/latex]:
[latex]\left(x+1\right)\left(3x-2\right)[/latex]
Answer
[latex]\left(x+1\right)\left(3x-2\right)[/latex]
Try It
Factor [latex]5x^{2}+5x–7x–7[/latex].
The following video presents another example of factoring by grouping when one of the GCF is negative.
Sometimes, we will encounter polynomials that, despite our best efforts, cannot be factored into the product of two binomials.
Example
Factor [latex]7x^{2}–21x+5x–5[/latex].
Solution
Group terms into pairs:
[latex]\left(7x^{2}–21x\right)+\left(5x–5\right)[/latex]
Factor the common factor [latex]7x[/latex] out of the first group, and the common factor [latex]5[/latex] out of the second group:
[latex]7x\left(x-3\right)+5\left(x-1\right)[/latex]
The two groups [latex]7x\left(x–3\right)[/latex] and [latex]5\left(x–1\right)[/latex] do not have any common factors, other than 1, so this polynomial cannot be factored any further.
[latex]7x\left(x–3\right)+5\left(x–1\right)[/latex]
Answer
Cannot be factored
In the example above, each pair can be factored, but then there is no common factor between the pairs! This means that the polynomial cannot be factored. In this case, the polynomial is said to be prime.
Try It
1. Factor [latex]-2x^{2}–2x+5x+5[/latex].
2. Factor [latex]6x^2+3x-8x-4[/latex]
3. Factor [latex]x^2-3x+x-3[/latex]
In the next section, we will see how factoring by grouping can be used to factor a trinomial.
