Learning Outcome
- Apply an algorithm to rewrite a trinomial as a four term polynomial and factor
- Use factoring by grouping to factor a trinomial
- Factor trinomials of the form [latex]a{x}^{2}+bx+c[/latex]
In the last section, we showed you how to factor polynomials with four terms by grouping. Trinomials of the form [latex]a{x}^{2}+bx+c[/latex] are slightly more complicated to factor. For trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[/latex] can be rewritten as [latex]\left(2x+3\right)\left(x+1\right)[/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[/latex] and then factor each portion of the expression to obtain [latex]2x\left(x+1\right)+3\left(x+1\right)[/latex]. We then pull out the GCF of [latex]\left(x+1\right)[/latex] to find the factored expression.
Below is a summary of the steps we will use followed by an example demonstrating how to use the step.
Factoring Trinomials in the form [latex]ax^{2}+bx+c[/latex]
To factor a trinomial in the form [latex]ax^{2}+bx+c[/latex], find two integers, r and s, whose sum is b and whose product is ac.
[latex]\begin{array}{l}r\cdot{s}=a\cdot{c}\\r+s=b\end{array}[/latex]
Rewrite the trinomial as [latex]ax^{2}+rx+sx+c[/latex] and then use grouping and the distributive property to factor the polynomial.
The first step in this process is to figure out what two numbers to use to re-write the x term as the sum of two new terms. Making a table to keep track of your work is helpful. We are looking for two numbers with a product of [latex]6[/latex] and a sum of [latex]5[/latex]
| Factors of [latex]2\cdot3=6[/latex] | Sum of Factors |
|---|---|
| [latex]1,6[/latex] | [latex]7[/latex] |
| [latex]-1,-6[/latex] | [latex]-7[/latex] |
| [latex]2,3[/latex] | [latex]5[/latex] |
| [latex]-2,-3[/latex] | [latex]-5[/latex] |
The pair [latex]p=2 \text{ and }q=3[/latex] will give the correct x term, so we will rewrite it using the new factors:
[latex]2{x}^{2}+5x+3=2x^2+2x+3x+3[/latex]
Now we can group the polynomial into two binomials.
[latex]2x^2+2x+3x+3=(2x^2+2x)+(3x+3)[/latex]
Identify the GCF of each binomial.
[latex]2x[/latex] is the GCF of [latex](2x^2+2x)[/latex] and [latex]3[/latex] is the GCF of [latex](3x+3)[/latex]. Use this to rewrite the polynomial:
[latex](2x^2+2x)+(3x+3)=2x(x+1)+3(x+1)[/latex]
Note how we leave the signs in the binomials and the addition that joins them. Be careful with signs when you factor out the GCF. The GCF of our new polynomial is [latex](x+1)[/latex]. We factor this out as well:
[latex]2x(x+1)+3(x+1)=(x+1)(2x+3)[/latex].
Sometimes it helps visually to write the polynomial as [latex](x+1)2x+(x+1)3[/latex] before you factor out the GCF. This is purely a matter of preference. Multiplication is commutative, so order does not matter.
Now let’s see how this strategy works for factoring [latex]6z^{2}+11z+4[/latex].
In this trinomial, [latex]a=6[/latex], [latex]b=11[/latex], and [latex]c=4[/latex]. According to the strategy, you need to find two factors, r and s, whose sum is [latex]b=11[/latex] and whose product is [latex]a\cdot{c}=6\cdot4=24[/latex]. You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since ac is positive and b is positive, you can be certain that the two factors you’re looking for are also positive numbers.)
| Factors whose product is [latex]24[/latex] | Sum of the factors |
|---|---|
| [latex]1\cdot24=24[/latex] | [latex]1+24=25[/latex] |
| [latex]2\cdot12=24[/latex] | [latex]2+12=14[/latex] |
| [latex]3\cdot8=24[/latex] | [latex]3+8=11[/latex] |
| [latex]4\cdot6=24[/latex] | [latex]4+6=10[/latex] |
There is only one combination where the product is [latex]24[/latex] and the sum is [latex]11[/latex], and that is when [latex]r=3[/latex], and [latex]s=8[/latex]. Let’s use these values to factor the original trinomial.
Example
Factor [latex]6z^{2}+11z+4[/latex].
Now, let’s look at an example where c is negative.
Example
Factor [latex]5{x}^{2}+7x - 6[/latex] by grouping.
We can summarize our process in the following way:
How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[/latex], factor by grouping
- List factors of [latex]ac[/latex].
- Find [latex]p[/latex] and [latex]q[/latex], a pair of factors of [latex]ac[/latex] with a sum of [latex]b[/latex].
- Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[/latex].
- Pull out the GCF of [latex]a{x}^{2}+px[/latex].
- Pull out the GCF of [latex]qx+c[/latex].
- Factor out the GCF of the expression.
Try It
In the following video, we present another example of factoring a trinomial using grouping. In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.
Using this strategy, factoring trinomials becomes quick and kind of fun once you get the idea. Give the next examples a try on your own before you look at the solution.
We will show two more examples so you can become acquainted with the variety of possible outcomes for factoring this type of trinomial.
Example
Factor [latex]2{x}^{2}+9x+9[/latex].
Here is an example for you to try where the c is negative.
Example
Factor [latex]6{x}^{2}+x - 1[/latex].
Prime Trinomials
Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[/latex], for instance. Can you think of two integers whose sum is [latex]b=35[/latex] and whose product is [latex]a\cdot{c}=2\cdot7=14[/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial. For our last example, you will see that sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.
Example
Factor [latex]7x^{2}-16x–5[/latex].
Factoring out a GCF
Now that you are familiar with the grouping method for factoring, we will now start to look at some strategies that are useful for making factoring easier. Your first step when factoring should always be to look for common factors for the three terms. Consider the examples below.
| Trinomial | Factor out Common Factor | Factored |
|---|---|---|
| [latex]2x^{2}+10x+12[/latex] | [latex]2(x^{2}+5x+6)[/latex] | [latex]2\left(x+2\right)\left(x+3\right)[/latex] |
| [latex]−5a^{2}−15a−10[/latex] | [latex]−5(a^{2}+3a+2)[/latex] | [latex]−5\left(a+2\right)\left(a+1\right)[/latex] |
| [latex]c^{3}–8c^{2}+15c[/latex] | [latex]c\left(c^{2}–8c+15\right)[/latex] | [latex]c\left(c–5\right)\left(c–3\right)[/latex] |
| [latex]y^{4}–9y^{3}–10y^{2}[/latex] | [latex]y^{2}\left(y^{2}–9y–10\right)[/latex] | [latex]y^{2}\left(y–10\right)\left(y+1\right)[/latex] |
Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.
Example
Factor [latex]3x^{3}–3x^{2}–90x[/latex].
The general form of trinomials with a leading coefficient of a is [latex]ax^{2}+bx+c[/latex]. Sometimes the factor of a can be factored as you saw above; this happens when a can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an [latex]x^{2}[/latex] term, instead of an [latex]ax^{2}[/latex] term.
Factoring out a Negative
In some situations, a is negative, as in [latex]−4h^{2}+11h+3[/latex]. It often makes sense to factor out [latex]−1[/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[/latex] from negative to positive, making the remaining trinomial easier to factor.
Example
Factor [latex]−4h^{2}+11h+3[/latex]
Note that the answer above can also be written as [latex]\left(−h+3\right)\left(4h+1\right)[/latex] or [latex]\left(h–3\right)\left(−4h–1\right)[/latex] if you multiply [latex]−1[/latex] times one of the other factors.
In the following video example, we will factor a trinomial whose leading term is negative. You will see how, by factoring out the negative sign, factoring the trinomial becomes easier.
