6.2 Factor Trinomials - Intermediate Algebra 2e | OpenStax (2024)

Factor Trinomials of the Form $x^2+bx+c$

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication. To factor the trinomial means to start with the product, and end with the factors.

To figure out how we would factor a trinomial of the form $x^2+bx+c,$ such as $x^2+5x+6$ and factor it to $\left(x+2\right)\left(x+3\right),$ let’s start with two general binomials of the form $\left(x+m\right)$ and $\left(x+n\right).$

Foil to find the product.
Factor the GCF from the middle terms.
Our trinomial is of the form $x^2+bx+c.$

This tells us that to factor a trinomial of the form $x^2+bx+c,$ we need two factors $\left(x+m\right)$ and $\left(x+n\right)$ where the two numbers m and n multiply to c and add to b.

Let’s summarize the steps we used to find the factors.

In the first example, all terms in the trinomial were positive. What happens when there are negative terms? Well, it depends which term is negative. Let’s look first at trinomials with only the middle term negative.

How do you get a positive product and a negative sum? We use two negative numbers.

Now, what if the last term in the trinomial is negative? Think about FOIL. The last term is the product of the last terms in the two binomials. A negative product results from multiplying two numbers with opposite signs. You have to be very careful to choose factors to make sure you get the correct sign for the middle term, too.

How do you get a negative product and a positive sum? We use one positive and one negative number.

When we factor trinomials, we must have the terms written in descending order—in order from highest degree to lowest degree.

Sometimes you’ll need to factor trinomials of the form $x^2+bxy+c{y}^2$ with two variables, such as $x^2+12xy+36y^2.$ The first term, $x^2,$ is the product of the first terms of the binomial factors, $x·x.$ The $y^2$ in the last term means that the second terms of the binomial factors must each contain y. To get the coefficients b and c, you use the same process summarized in How To Factor trinomials.

Some trinomials are prime. The only way to be certain a trinomial is prime is to list all the possibilities and show that none of them work.

Let’s summarize the method we just developed to factor trinomials of the form $x^2+bx+c.$

Factor Trinomials of the form ax² + bx + c using Trial and Error

Our next step is to factor trinomials whose leading coefficient is not 1, trinomials of the form $a{x}^2+bx+c.$

Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods we’ve used so far. Let’s do an example to see how this works.

What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

Let’s factor the trinomial $3x^2+5x+2.$

From our earlier work, we expect this will factor into two binomials.

$$\begin{array}{}\\ \\ \hfill 3x^2+5x+2\hfill \\ \hfill \left(\right)\left(\right)\hfill \end{array}$$

We know the first terms of the binomial factors will multiply to give us $3x^2.$ The only factors of $3x^2$ are $1x,3x.$ We can place them in the binomials.

Check: Does $1x·3x=3x^2?$

We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1, 2. But we now have two cases to consider as it will make a difference if we write 1, 2 or 2, 1.

Which factors are correct? To decide that, we multiply the inner and outer terms.

Since the middle term of the trinomial is $5x,$ the factors in the first case will work. Let’s use FOIL to check.

$$\begin{array}{c}\hfill \left(x+1\right)\left(3x+2\right)\hfill \\ \hfill 3x^2+2x+3x+2\hfill \\ \hfill 3x^2+5x+2✓\hfill \end{array}$$

Our result of the factoring is:

$$\begin{array}{}\\ \hfill 3x^2+5x+2\hfill \\ \hfill \left(x+1\right)\left(3x+2\right)\hfill \end{array}$$

Remember, when the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

Example 6.16

Factor completely using trial and error: $6b^2-13b+5.$

Solution

The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since the last term, 5, is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors.

Consider all the combinations of factors.

$6b^2-13b+5$
Possible factors	Product
$\left(b-1\right)\left(6b-5\right)$	$6b^2-11b+5$
$\left(b-5\right)\left(6b-1\right)$	$6b^2-31b+5$
$\left(2b-1\right)\left(3b-5\right)$	$6b^2-13b+5^{*}$
$\left(2b-5\right)\left(3b-1\right)$	$6b^2-17b+5$

The correct factors are those whose product is the original trinomial.	$\left(2b-1\right)\left(3b-5\right)$
Check by multiplying: $\begin{array}{}\\ \\ \hfill \left(2b-1\right)\left(3b-5\right)\hfill \\ \hfill 6b^2-10b-3b+5\hfill \\ \hfill 6b^2-13b+5✓\hfill \end{array}$

When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

Example 6.17

Factor completely using trial and error: $18x^2-37xy+15y^2.$

Solution

The trinomial is already in descending order.
Find the factors of the first term.
Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative factors.

Consider all the combinations of factors.

The correct factors are those whose product is the original trinomial.	$\left(2x-3y)(9x-5y\right)$
Check by multiplying: $\begin{array}{}\\ \\ \hfill \left(2x-3y)(9x-5y\right)\hfill \\ \hfill 18x^2-10xy-27xy+15y^2\hfill \\ \hfill 18x^2-37xy+15y^2✓\hfill \end{array}$

Don’t forget to look for a GCF first and remember if the leading coefficient is negative, so is the GCF.

Example 6.18

Factor completely using trial and error: $\mathrm{-10}{y}^4-55y^3-60y^2.$

Solution

Notice the greatest common factor, so factor it first.
Factor the trinomial.

Consider all the combinations.

The correct factors are those whose product is the original trinomial. Remember to include the factor $\text{−}5y^2.$	$\text{−}5y^2\left(y+4\right)\left(2y+3\right)$
Check by multiplying: $\begin{array}{}\\ \\ \hfill \text{−}5y^2\left(y+4\right)\left(2y+3\right)\hfill \\ \hfill \text{−}5y^2\left(2y^2+8y+3y+12\right)\hfill \\ \hfill \text{−}10y^4-55y^3-60y^2✓\hfill \end{array}$

Factor Trinomials of the Form $a{x}^2+bx+c$ using the “ac” Method

Another way to factor trinomials of the form $a{x}^2+bx+c$ is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!

The “ac” method is summarized here.

Don’t forget to look for a common factor!

Example 6.20

Factor using the ‘ac’ method: $10y^2-55y+70.$

Solution

Is there a greatest common factor?
Yes. The GCF is 5.
Factor it.
The trinomial inside the parentheses has a leading coefficient that is not 1.
Find the product $ac.$	$ac=28$
Find two numbers that multiply to $ac$	$(\mathrm{-4})(\mathrm{-7})=28$
and add to b.	$\mathrm{-4}+(\mathrm{-7})=\mathrm{-11}$
Split the middle term.
Factor the trinomial by grouping.
Check by multiplying all three factors. $\begin{array}{c}\hfill 5\left(y-2\right)\left(2y-7\right)\hfill \\ \hfill 5\left(2y^2-7y-4y+14\right)\hfill \\ \hfill 5\left(2y^2-11y+14\right)\hfill \\ \hfill 10y^2-55y+70✓\hfill \end{array}$

Factor Using Substitution

Sometimes a trinomial does not appear to be in the $a{x}^2+bx+c$ form. However, we can often make a thoughtful substitution that will allow us to make it fit the $a{x}^2+bx+c$ form. This is called factoring by substitution. It is standard to use u for the substitution.

In the $a{x}^2+bx+c,$ the middle term has a variable, x, and its square, $x^2,$ is the variable part of the first term. Look for this relationship as you try to find a substitution.

Example 6.21

Factor by substitution: $x^4-4x^2-5.$

Solution

The variable part of the middle term is $x^2$ and its square, $x^4,$ is the variable part of the first term. (We know ${\left(x^2\right)}^2=x^4).$ If we let $u=x^2,$ we can put our trinomial in the $a{x}^2+bx+c$ form we need to factor it.

Rewrite the trinomial to prepare for the substitution.
Let $u=x^2$ and substitute.
Factor the trinomial.
Replace u with $x^2.$
Check: $\begin{array}{c}\hfill \left(x^2+1\right)\left(x^2-5\right)\hfill \\ \hfill x^4-5x^2+x^2-5\hfill \\ \hfill x^4-4x^2-5✓\hfill \end{array}$

Sometimes the expression to be substituted is not a monomial.

Example 6.22

Factor by substitution: ${\left(x-2\right)}^2+7\left(x-2\right)+12$

Solution

The binomial in the middle term, $\left(x-2\right)$ is squared in the first term. If we let $u=x-2$ and substitute, our trinomial will be in $a{x}^2+bx+c$ form.

Rewrite the trinomial to prepare for the substitution.
Let $u=x-2$ and substitute.
Factor the trinomial.
Replace u with $x-2.$
Simplify inside the parentheses.

This could also be factored by first multiplying out the ${\left(x-2\right)}^2$ and the $7\left(x-2\right)$ and then combining like terms and then factoring. Most students prefer the substitution method.