There are many ways to factor, but it’s always good to know the shortcuts that will make the process that much faster. If you’re working on factoring, you might want to consider a couple of the ideas below for factoring by grouping. Get some extra help with factoring with an online class.

## Factor By Grouping Using the AC Formula

A trinomial can easily be broken down into manageable parts. They usually follow a pattern like the form below:

ax^{2} + bx + c

Usually, the three other letters – a, b, and c – would be replaced with numerical values. The simplest polynomials usually involve a having the value of 1. If a is equal to 1, the rest of the problem is quick and easy to solve. However, what if a is not equal to 1? What if a is a larger number? Here’s an example:

3x^{2} + 8x + 4

There are four simple steps to solving this equation that won’t have you spending several minutes to solve each problem. Let’s solve this problem using those steps, and they will be outlined in more detail after the example.

**ac = d**: 3 · 4 = 12

**factors of d; factors equal to b:** 1, 2, 3, 4, 6, 12; 2 + 6 = 8

**ungroup:** 3x^{2} + 2x + 6x + 4

**group:** (3x^{2} + 2x) + (6x + 4)

x(3x + 2) + 2(3x + 2)

(x + 2)(3x + 2)

Obviously, without knowing what you’re doing, the solving above looks like some strange mixture of numbers and letters. However, the four steps are simple and are outlined below.

- Begin by deciding the product of a and c. In the case of the above problem, this was 3 and 4. 3 multiplied by 4 gives you 12, which was called d.
- Factor 12, and decide what factors will give you b. 12 can be factored into several numbers, but only two of them when added together will give you b, which was 8. These factors are 2 and 6.
- Using your new factors, you ungroup the middle term. This was what created the new problem listed above beside “ungroup.”
- This step has several sub-steps, but they are all pretty easy as well. Once you have the new middle terms, you group the problem into two pairs as shown above: (3x
^{2}+ 2x) + (6x + 4). You then remove the common factors from the two pairs. In the case of (3x^{2}+ 2x), this common factor was a single x. (6x + 4) shared a 2 in common. This then gave you a new equation: x(3x + 2) + 2(3x + 2). As you can see, the first pair is obvious because it can be factored out – (3x + 2). You then take the remaining factors outside the parenthesis to create the second binomial – (x + 2).

Here’s another problem using the steps listed above:

**5x ^{2} + 18x + 9**

5 · 9 = 45

45: 1, 3, 5, 9, 15, 45

3 + 15 = 18

5x^{2} + 3x + 15x + 9

(5x^{2} + 3x) + (15x + 9)

x(5x + 3) + 3(5x + 3)

(x + 3)(5x + 3)

You can check your answer using the FOIL method to make certain it is in fact the factoring of the polynomial:

F: x · 5x = 5x^{2}

O: x · 3 = 3x

I: 3 · 5x = 15x

L: 3 · 3 = 9

5x^{2} + 3x + 15x + 9

5x^{2} + 18x + 9

If you end up with the original problem in the end, you have successfully factored your problem using the AC formula. Get straight to the point with Algebra I in an online course.

## Using a Similar Method to Solve Polynomials with More Than Three Terms

Algebra I is complicated in the sense that trinomials and binomials aren’t the only kinds of polynomials you’re likely to run into. You will more than likely end up with polynomials that have four terms. You could also run into polynomials with more than just x as variables. Examples of both will be given here along with how to solve them using grouping. The method is similar to the method you learned above with a couple of the steps removed. Learn algebra from beginning to end with an online class.

Here is an example of a polynomial with more than three terms and only x as a variable:

x^{3} + 2x^{2} + 8x + 16

As you can see, the AC formula won’t work so well with the first three steps as there is a fourth term now that throws a wrench into it all. However, you can use a modified version of the fourth step, which is shown below:

**x ^{3} + 2x^{2} + 8x + 16**

(x^{3 + 2x2) + (8x + 16)}

^{ x2(x + 2) + 8(x + 2)}

^{ (x + 2)(x2 + 8)}

Rather than finding the factors of a term that add up to be a middle term, you can easily factor out a binomial that is created by grouping terms together. Because you already have four terms, there’s no need to multiply anything and find factors to ungroup a middle term to create four terms. You simply group the four terms together, factor out common factors from each of the two groups, then factor out the common binomial to get the second binomial. Learn great strategies for solving algebraic equations with an online course. Now, let’s make things trickier by adding a second variable into the mix:

xy – 4y + 3x – 12

On the bright side, you no longer have exponents to worry about. However, you now have a problem that includes two different variables, and until now, your binomials always have the same variable. Not to worry, however. The same modified step four from the AC formula can be used to solve this problem. Look how you can factor the problem above:

**xy – 4y + 3x – 12**

(xy – 4y) + (3x – 12)

y(x – 4) + 3(x – 4)

(x – 4)(y + 3)

There is no rule that says binomials have to include the same variable in them. As long as you check your answer using FOIL and end up with the original problem, you have likely factored your polynomial. Use FOIL on the problem above to make certain you did your work correctly:

**(x – 4)(y + 3)**

**F:** x · y = xy

**O:** x · 3 = 3x

**I:** -4 · y = -4y

**L:** -4 · 3 = -12

xy + 3x – 4y – 12

**rearrange the terms:** xy – 4y + 3x – 12

The polynomial matches the original polynomial that was factored. You now have three successful methods to factor problems by grouping. It will save you time and energy, and you won’t have as much scrap paper lying around on your desk.