Algebra can prove to be a difficult subject at times, and it can be even more difficult if you don’t practice the skills you once knew. If you’d like to revisit all of Algebra I, try taking an online course. If you just need a refresher in factoring completely, try these steps below.

Factoring completely involves combining three basic techniques involved in factoring – finding the greatest common factor, the difference between two squares, and the use of the trinomial. Before you begin with factoring completely, you might want to refresh yourself on these three important basics.

## Greatest Common Factor

Before you can find the greatest common factor of a trinomial, you’re going to need to know the greatest common factor for the three terms in the trinomial. Get straight to the point with Algebra I by taking an online class. Here’s an example problem of greatest common factor:

4x^{3} + 64x^{2} + 16x

The first thing you’re going to want to do is separate the terms from the rest of the problem. If you need to work out what the greatest common factor of the three terms on paper, list them on paper and find out the factors of all the three terms.

4 (1, 2, 4)

64 (1, 2, 4, 8, 16, 32, 64)

16 (1, 2, 4, 8, 16)

As you can see, the greatest common factor would be 2x as each term can be divided evenly by 2, and each term has an x that can be divided from it. You will then move onto the next step by writing the greatest common factor to the left of a set of parentheses:

2x( )

You will then divide each term by 2x and write the new trinomial in the parentheses:

4x^{3}/2x = 2x^{2}

64x^{2}/2x = 32x

16x/2x = 8

2x(2x^{2} + 32x + 8)

## Difference Between Two Squares

Some polynomials will include only two terms. This is usually because the third term that would be between them is cancelled out when doing the arithmetic involved in factoring. Here is an example problem:

(x – 4)(x + 4)

If you’re familiar with the FOIL method, you will know exactly what to do next. However, for those of you that need a review, here it is:

**F**: This letter stands for “firsts”, meaning you multiply the first terms. Multiply the first term in the first parentheses with the first term in the second parentheses. In the case of the above problem, you would multiply x by x. This would give you x^{2}.**O**: This letter means “outers”, meaning you multiply the two outer terms of the equation together. Multiply the first term in the first parentheses with the second term in the second parentheses. This would mean you multiply x by 4 in the above problem, giving you 4x.**I**: This letters stands for the “inners”, meaning you multiply the two inner terms of the equation. Multiply the second term in the first parentheses with the first term in the second parentheses. Keep in mind that the sign in front of these numbers is very important when FOILing. In the above problem, you would be multiplying -4 by x, and that would give you -4x.**L**: This means “lasts”, so you will now be multiplying the two last terms. Multiply the last term of the first parentheses with the last term in the second parentheses. This would mean multiplying -4 by 4, which would give you -16.

So, if you use the FOIL method on the problem above, you end up with a new equation that looks something like this:

x^{2} + 4x – 4x – 16

x^{2} ~~+ 4x – 4x~~ – 16

x^{2} – 16

The middle terms cancel out because 4x – 4x = 0. Therefore, there is no center term when dealing with the difference of two squares. Now, let’s take a problem and create the difference of two squares from it. Learn more about beginning algebraic concepts with an online course.

x^{4} – 25

The quickest way to figure out the answer to this problem is to take the square root of each term and use those new terms to create the parentheses:

√x^{4} = x^{2}

√25 = 5

(x^{2} + 5)(x^{2} – 5)

Looks simple enough, right? Well, before you move on to the next problem, you’ll want to make certain you were right by using FOIL to check your answer:

(x^{2} + 5)(x^{2} – 5)

x^{4} – 5x^{2} + 5x^{2} – 25

x^{4} ~~– 5x~~ – 25^{2} + 5x^{2}

x^{4} – 25

If you end up with the same equation that you started with, your job is now done.

## Factoring a Trinomial

Factoring a trinomial is very similar to finding the answer to a difference of two squares. First, you start with your trinomial:

x^{2} + 7x + 12

Like finding the greatest common factor, you’re going to need know all the different factors of 12 to know what terms you will use in the parentheses. As the middle term is 7, you’re going to want to find the factors that will add up to give you 7. This would be 3 and 4:

(x + 4)(x + 3)

To be certain your answer is correct, use the FOIL method to check your answer:

(x + 4)(x + 3)

F: x · x = x^{2}

O: x · 3 = 3x

I: 4 · x = 4x

L: 4 · 3 = 12

x^{2} + 3x + 4x + 12

x^{2} + 7x + 12

Again, if your answer matches the original problem, you have done it correctly. If you need more help with polynomials and factoring, try an online course for more help.

## Factoring Completely

The process of factoring completely is important when you come across problems that require extra steps to be completely factored. Again, factoring completely involves combining the three different techniques that were just discussed. Not every basic technique will be used in every problem, but the key to factoring completely is to check each of the basic techniques to be certain the problem is as complete as it can be. Here is an example problem:

12x^{4} – 3x^{2} – 54

The first step is to use the technique of factoring out the greatest common factor. In the case of the problem above, it would be 3:

3(4x^{4} – x^{2} – 18)

The second step would be to factor a trinomial into two binomials as there is no difference of squares yet.

3(4x^{2} – 9)(x^{2} + 2)

The first binomial is an obvious difference of squares so the final step would be to factor it. Remember to use the square roots of each term of the binomial to come up with the terms to be used in the parentheses:

4x^{2} – 9

√4x^{2} = 2x

√9 = 3

(2x – 3)(2x + 3)

You would then return to the problem created in the second step using the two binomials created in the third step instead of the first binomial:

3(2x – 3)(2x +3)(x^{2} + 2)

Again, before deciding if you’re done, you should always check your answer. Obviously, checking your answer with a problem like this can be a bit more difficult. First, you begin with FOIL. You can start with any of the three binomials, but it might be easier to start with the difference of two squares:

(2x – 3)(2x + 3)

F: 2x · 2x = 4x^{2}

O: 2x · 3 = 6x

I: -3 · 2x = -6x

L: -3 · 3

4x^{2} + 6x – 6x – 9

4x^{2} ~~+ 6x – 6x~~ – 9

4x^{2} – 9

You would then take this binomial and return it to the original problem:

3(4x^{2} – 9)(x^{2} + 2)

You would then FOIL the two remaining binomials to get the original trinomial with the 3 factored out:

(4x^{2} – 9)(x^{2} + 2)

F: 4x^{2} · x^{2} = 4x^{4}

O: 4x^{2} · 2 = 8x^{2}

I: -9 · x^{2} = -9x^{2}

L: -9 · 2 = -18

4x^{4} + 8x^{2} – 9x^{2} – 18

4x^{4} – x^{2} – 18

You would then take this trinomial and return it to the original problem just before this step:

3(4x^{4} – x^{2} – 18)

You would then multiply each term by the factored-out 3:

12x^{4} – 3x^{2} – 54

If you end up with the original problem, you have done it correctly. You have now factored a problem completely and then checked your answer.