# The Grouping Number Method

This is an excerpt from a previous post, but if I want to introduce someone to the grouping number method, they don’t need all of the commentary about the AC method.

Are you looking for a nice and intuitive method for factoring quadratics when the leading coefficient isn’t one? Most people were probably taught to break these these things down by trial and error. That isn’t awful once you’ve developed some intuition, but if you haven’t, trial and error quickly becomes tedious. What should we start with instead? The grouping number method works well, especially if you’re just starting out.

To set up the Grouping Number Method, let’s think about what happens in the following polynomial multiplication.

$(4x-1)(2x-3)$

$=2x(4x-1)-3(4x-1)$

$=8x^2 -2x -12x +3$

$=8x^2 -14x +3$

Normally, you’d just skip the second step, but it’s important to realize that what you’re really doing is using the distributive property twice. Additionally, let’s think about the -12 and the -2. It’s clear that they add up to the final coefficient of x, but it’s also true that their product (-12)(-2) = 24 is the same as the product of the leading coefficient and the constant term, (8)(3). Notice this has to be true. In either case, it’s the product of 4, -1, 2 and -3, merely in a different order.

So, how do we go in the other direction? That is, how do we factor instead of multiply? We’ll demonstrate the grouping number method on the polynomial $8x^2 - 14x + 3$.

The grouping number of a quadratic polynomial is the leading coefficient multiplied by the constant term.

So, in this case, it would be $8 times 3 = 24$.

Next, you want to look for two factors of your grouping number that add up to the coefficient of $x$. In this case, $(-12)(-2) = 24$ and $(-12)+(-2) = -14$. We use these two numbers to rewrite the middle term to get the following.

$8x^2 - 14x + 3 = 8x^2 - 2x - 12x + 3$.

In other words, to rewrite the polynomial in this way, you find two factors of your grouping number that add up to the coefficient of $x$ and use those to break down the middle term.

Why rewrite the polynomial in this way? Because it sets us up perfectly to factor by grouping.

Factoring by grouping is a technique for factoring a polynomial with four terms. In a nutshell, to factor by grouping, you remove the greatest common factor from the first two terms, then remove the greatest common factor from the last two terms. If the resulting binomial factors are the same you can factor this out to get the product of two binomials. Notice, that’s using the distributive property twice, just in the opposite direction as before.

Therefore, to finish factoring the polynomial we can factor by grouping.

$8x^2 - 14x + 3$

$= 8x^2 - 2x - 12x + 3$

$= 2x(4x - 1) - 3(4x - 1)$

$= (4x - 1)(2x - 3)$

Notice this exactly reverses the steps of the multiplication with which we started.

In short, as long as you have no common factors, the quadratic polynomials where the grouping number can be factored into two numbers that add up to the middle coefficient are exactly the ones that can be factored by grouping. All the others are prime.