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How to factor when the leading coefficient isn’t one.

In the previous section we developed the method of factoring by grouping and the section before that we developed a method of factoring coefficients to factor quadratics with a leading coefficient of one. We can use the same method of factoring coefficients on polynomials that have a leading coefficient that isn’t one, but it gets much more messy and difficult to use. Instead there is a generally easier way to tackle this method that is actually a hybrid of the factoring coefficients and factor by grouping technique, called the “AC Method”.
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Let’s consider the quadratic in its expanded form;

Remember that the goal for factor by grouping is to arrange things so that the same factor appears in all the “groupings” after factoring out the GCF. The first step to figuring out the right “groupings” here is to split the middle term into two pieces and group the first term with one of those pieces, and the last term with another of those pieces. Thus we want to (somehow) do the following:

In order for the above to be valid we obviously need that $b = d + k$ (since we replaced $bx$ with $dx + kx$). But even more difficult; in order for the above to help we need to somehow arrange things so that once we factor out the GCF in the first and second groups, the leftover is the same thing. This seems like a crazy coincidence to rig, but it turns out that doing so it easier than it seems, if a little unintuitive.

#### AC Method: Rigging the System

In order to rig things so that we get everything we want, we will perform factoring by coefficients; but not on the coefficients we were given. Instead we will do it by replacing the $c$ term in our original polynomial with $a \cdot c$, then we can use the coefficients factoring result to split our quadratic into 4 terms and factor the result using grouping. This is easiest to see as an example.

So, using the AC method breaks down to the following steps:

(a)
First multiply the leading coefficient with the constant term; that is multiply $a \cdot c$. (In our example above: $6 \cdot 5$)
(b)
Find two numbers that multiply to the number from the previous step and add to the middle terms coefficient. (In our example: two numbers that multiply to $30$ but add to $17$. We settled on $2$ and $15$.) Note: One or both of these numbers might be negative!
(c)
Split the middle term of the original polynomial into two terms, the coefficients of which are the factors you found in the previous step. (In our example: We settled on $2$ and $15$ so we replaced the middle term: $17x$, with $2x + 15x$.)
(d)
The modified polynomial now has four terms, so we can apply Factor by Grouping. This should result in our factored answer. (In our example: Apply factor by grouping to $6x^2 + 15x + 2x + 5$ to ultimately get $(3x + 1)(2x + 5)$.)