Factor Trinomials With Leading Coefficient Greater Than 1

Learning Objectives

Factor trinomials of the form $ax^2+bx+c$
Recognize where to place negative signs when factoring a trinomial
Recognize when a polynomial is a difference of squares, and how it would factor as the product of two binomials

Because we couldn’t get enough factoring, we thought we would give an encore.

Encore

In this section we will continue to factor trinomials so you can become proficient with using the techniques presented in the last section. The goal is for you to be comfortable with recognizing where to place negative signs, and whether a trinomial can even be factored. Additionally, we will introduce one special case to look out for at the end of this page.

Not all trinomials look like $x^{2}+5x+6$ , where the coefficient in front of the $x^{2}$ term is 1. In these cases, your first step should be to look for common factors for the three terms.

Trinomial	Factor out Common Factor	Factored
$2x^{2}+10x+12$	$2(x^{2}+5x+6)$	$2\left(x+2\right)\left(x+3\right)$
$−5a^{2}−15a−10$	$−5(a^{2}+3a+2)$	$−5\left(a+2\right)\left(a+1\right)$
$c^{3}–8c^{2}+15c$	$c\left(c^{2}–8c+15\right)$	$c\left(c–5\right)\left(c–3\right)$
$y^{4}–9y^{3}–10y^{2}$	$y^{2}\left(y^{2}–9y–10\right)$	$y^{2}\left(y–10\right)\left(y+1\right)$

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 $3x^{3}–3x^{2}–90x$ .

Show Solution

Since 3 is a common factor for the three terms, factor out the 3.

$3\left(x^{3}–x^{2}–30x\right)$

x is also a common factor, so factor out x.

$3x\left(x^{2}–x–30\right)$

Now you can factor the trinomial $x^{2}–x–30$ . To find r and s, identify two numbers whose product is $−30$ and whose sum is $−1$ .

The pair of factors is $−6$ and $5$ . So replace $–x$ with $−6x+5x$ .

$3x\left(x^{2}–6x+5x–30\right)$

Use grouping to consider the terms in pairs.

$3x\left[\left(x^{2}–6x\right)+\left(5x–30\right)\right]$

Factor x out of the first group and factor 5 out of the second group.

$3x\left[\left(x\left(x–6\right)\right)+5\left(x–6\right)\right]$

Then factor out $x–6$ .

$3x\left(x–6\right)\left(x+5\right)$

Answer

$3x\left(x–6\right)\left(x+5\right)$

The following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF. We use the shortcut method instead of factoring by grouping.

The general form of trinomials with a leading coefficient of a is $ax^{2}+bx+c$ . 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 $x^{2}$ term, instead of an $ax^{2}$ term.

However, if the coefficients of all three terms of a trinomial don’t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.

Factoring Trinomials in the form $ax^{2}+bx+c$

To factor a trinomial in the form $ax^{2}+bx+c$ , find two integers, r and s, whose sum is b and whose product is ac.

$\begin{array}{l}r\cdot{s}=a\cdot{c}\\r+s=b\end{array}$

Rewrite the trinomial as $ax^{2}+rx+sx+c$ and then use grouping and the distributive property to factor the polynomial.

This is almost the same as factoring trinomials in the form $x^{2}+bx+c$ , as in this form $a=1$ . Now you are looking for two factors whose product is $a\cdot{c}$ , and whose sum is b.

Let’s see how this strategy works by factoring $6z^{2}+11z+4$ .

In this trinomial, $a=6$ , $b=11$ , and $c=4$ . According to the strategy, you need to find two factors, r and s, whose sum is $b=11$ and whose product is $a\cdot{c}=6\cdot4=24$ . 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 24	Sum of the factors
$1\cdot24=24$	$1+24=25$
$2\cdot12=24$	$2+12=14$
$3\cdot8=24$	$3+8=11$
$4\cdot6=24$	$4+6=10$

There is only one combination where the product is 24 and the sum is 11, and that is when $r=3$ , and $s=8$ . Let’s use these values to factor the original trinomial.

Example

Factor $6z^{2}+11z+4$ .

Show Solution

Rewrite the middle term,

11 z

$11z$ , as

3 z + 8 z

$3z + 8z$ (from the chart above.)

$6z^{2}+3z+8z+4$

Group pairs. Use grouping to consider the terms in pairs.

$\left(6z^{2}+3z\right)+\left(8z+4\right)$

Factor 3z out of the first group and 4 out of the second group.

$3z\left(2z+1\right)+4\left(2z+1\right)$

Factor out $\left(2z+1\right)$ .

$\left(2z+1\right)\left(3z+4\right)$

Answer

$\left(2z+1\right)\left(3z+4\right)$

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.

Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial $2z^{2}+35z+7$ , for instance. Can you think of two integers whose sum is $b=35$ and whose product is $a\cdot{c}=2\cdot7=14$ ? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.

In some situations, a is negative, as in $−4h^{2}+11h+3$ . It often makes sense to factor out $−1$ as the first step in factoring, as doing so will change the sign of $ax^{2}$ from negative to positive, making the remaining trinomial easier to factor.

Example

Factor $−4h^{2}+11h+3$

Show Solution

Factor

- 1

$−1$ out of the trinomial. Notice that the signs of all three terms have changed.

$−1\left(4h^{2}–11h–3\right)$

To factor the trinomial, you need to figure out how to rewrite $−11h$ . The product of $rs=4\cdot−3=−12$ , and the sum of $rs=−11$ .

$r\cdot{s}=−12$	$r+s=−11$
$−12\cdot1=−12$	$−12+1=−11$
$−6\cdot2=−12$	$−6+2=−4$
$−4\cdot3=−12$	$−4+3=−1$

Rewrite the middle term $−11h$ as $−12h+1h$ .

$−1\left(4h^{2}–12h+1h–3\right)$

Group terms.

$−1\left[\left(4h^{2}–12h\right)+\left(1h–3\right)\right]$

Factor out 4h from the first pair. The second group cannot be factored further, but you can write it as $+1\left(h–3\right)$ since $+1\left(h–3\right)=\left(h–3\right)$ . This helps with factoring in the next step.

$−1\left[4h\left(h–3\right)+1\left(h–3\right)\right]$

Factor out a common factor of $\left(h–3\right)$ . Notice you are left with $\left(h–3\right)\left(4h+1\right)$ ; the $+1$ comes from the term $+1\left(h–3\right)$ in the previous step.

$−1\left[\left(h–3\right)\left(4h+1\right)\right]$

Answer

$−1\left(h–3\right)\left(4h+1\right)$

Note that the answer above can also be written as $\left(−h+3\right)\left(4h+1\right)$ or $\left(h–3\right)\left(−4h–1\right)$ if you multiply $−1$ times one of the other factors.

In the following video we present another example of factoring a trinomial in the form $-ax^2+bx+c$ using the grouping technique.

Difference of Squares

We would be remiss if we failed to introduce one more type of polynomial that can be factored. This polynomial can be factored into two binomials but has only two terms. Let’s start from the product of two binomials to see the pattern.

Given the product of two binomials: $\left(x-2\right)\left(x+2\right)$ , if we multiply them together, we lose the middle term that we are used to seeing as a result.

Multiply:

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

The polynomial $x^2-4$ is called a difference of squares because teach term can be written as something squared. A difference of squares will always factor in the following way:

Factor a Difference of Squares

Given $a^2-b^2$ , it’s factored form will be $\left(a+b\right)\left(a-b\right)$

Let’s factor $x^{2}–4$ by writing it as a trinomial, $x^{2}+0x–4$ . This is similar in format to the trinomials we have been factoring so far, so let’s use the same method.

Find the factors of $a\cdot{c}$ whose sum is b, in this case, 0:

Factors of $−4$	Sum of the factors
$1\cdot-4=−4$	$1-4=−3$
$2\cdot−2=−4$	$2-2=0$
$-1\cdot4=−4$	$-1+4=3$

2, and -2 have a sum of 0. You can use these to factor $x^{2}–4$ .

Example

Factor $x^{2}–4$ .

Show Solution

Rewrite

0 x

$0x$ as

- 2 x + 2 x

$−2x+2x$ .

$\begin{array}{l}x^{2}+0x-4\\x^{2}-2x+2x-4\end{array}$

Group pairs.

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

Factor x out of the first group. Factor 2 out of the second group.

$x\left(x–2\right)+2\left(x–2\right)$

Factor out $\left(x–2\right)$ .

$\left(x–2\right)\left(x+2\right)$

Answer

$\left(3x–2\right)\left(3x+2\right)$

Since order doesn’t matter with multiplication, the answer can also be written as $\left(x+2\right)\left(x–2\right)$ .

You can check the answer by multiplying $\left(x–2\right)\left(x+2\right)=x^{2}+2x–2x–4=x^{2}–4$ .

The following video show two more examples of factoring a difference of squares.

Summary

When a trinomial is in the form of $ax^{2}+bx+c$ , where a is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form $ax^{2}+bx+c$ , find two integers, r and s, whose sum is b and whose product is ac. Then rewrite the trinomial as $ax^{2}+rx+sx+c$ and use grouping and the distributive property to factor the polynomial.

When $ax^{2}$ is negative, you can factor $−1$ out of the whole trinomial before continuing.

A difference of squares $a^2-b^2$ will factor in this way: $\left(a+b\right)\left(a-b\right)$ .

Objective 3