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.

This is very similar to factoring trinomials in the form [latex]x^{2}+bx+c[/latex], except now you are looking for two factors whose product is [latex]a\cdot{c}[/latex], 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 [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]. 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.)

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

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 [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.

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.

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 we present another example of factoring a trinomial in the form [latex]-ax^2+bx+c[/latex] using the grouping technique.

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