6.4: Factoring Trinomials with Leading Coefficient other than 1

section 6.4 Learning Objectives

6.4: Factoring Trinomials with Leading Coefficient Other Than 1

Factor trinomials of the type ax² + bx + c, where a ≠ 1
Factor trinomials of the above type where the GCF must first be factored out

In Section 6.3, we factored trinomials with a leading coefficient of 1, or were able to factor out a common factor so that the leading coefficient became 1. It was this leading coefficient of 1 that allowed for the nice shortcut provided by the “Product and Sum Method.” However, if the leading coefficient is something other than 1 and the coefficients of all three terms of a trinomial don’t have a common factor (other than 1), then you will need to factor a trinomial with a leading 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 very similar to factoring trinomials in the form $x^{2}+bx+c$ , except now you are looking for two factors whose product is $a\cdot{c}$ , and whose sum is b. Additionally, we will find that there is not a simple shortcut like we had when the leading coefficient was 1, and will stick with the full grouping strategy.

Because our first step is to multiply the values for a and c, this technique is sometimes referred to as the “AC-Method.”

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 $a\cdot{c}=6\cdot4=24$ . Like before, 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 1

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

Show Solution

Rewrite the middle term,

$11z$ , as

$3z + 8z$ (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

$6z^2+11z+4=\left(2z+1\right)\left(3z+4\right)$

Example 2

Factor $10x^2-7x-6$

Show Solution

First we compute the product $a\cdot c$ where $a=10$ and $c=-6$ . We get

$a\cdot c=10(-6)=-60$

Next, we must find two integers that multiply to $-60$ and add to $-7$ (you can begin by listing out all pairs that multiply to $-60$ if you find this helpful). The integers will be $-12$ and $5$ .

Now we can rewrite the problem and factor by grouping.

$10x^2-7x-6$

$=10x^2-12x+5x-6$

$=\left(10x^2-12x\right)+\left(5x-6\right)$

$=2x\left(5x-6\right)+1\left(5x-6\right)$

$=\left(5x-6\right)\left(2x+1\right)$

Answer

$10x^2-7x-6=\left(5x-6\right)\left(2x+1\right)$

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 coefficient on the 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 3

Factor $−4h^{2}+11h+3$ .

Show Solution

Factor

$−1$ 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.

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

or simply

$-(h-3)(4h+1)$

Answer

$-4h^2+11h+3=−\left(h–3\right)\left(4h+1\right)$

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.

Similar to factoring out a $-1$ , we could encounter a more substantial GCF, which we see in our last example.

Example 4

Factor $16x^4-12x^3-10x^2$ .

Show Solution

Note that all three terms share a common factor of $2x^2$ . We factor this out first.

$2x^2(8x^2-6x-5)$

Since the resulting trinomial still has a leading coefficient other than 1, in this case $a=8$ , we must apply the strategy described in this section. We need two factors that multiply to $a\cdot c=8\cdot (-5)=-40$ while adding to $b=-6$ . Those two factors are $-10$ and $4$ .

Next, rewrite the middle term, $-6x$ , as $-10x+4x$ and use grouping.

$2x^2\left( 8x^2-10x+4x-5\right)$

$2x^2\left[ 2x(4x-5)+1(4x-5)\right]$

$2x^2 (4x-5)(2x+1)$

Answer

$16x^4-12x^3-10x^2=2x^2 (4x-5)(2x+1)$

Module 6: Factoring Polynomials