AC Method Factoring

"AC method factoring" is another method of factoring you can use when working with trinomials.  That is, adding something of the form ax² + bx + c.  This can be added to my list of methods of factoring that I posted about earlier, but is specifically used for trinomial factoring.  This is a simple method that will help you to solve problems faster, and it really isn't all that complicated.  However, as with any of the factoring methods, you have to learn and understand the steps to perform it.

This factoring method requires only a few steps, so I will go over them with you now, and then show you an example or two.  To start factoring equations, we need to have a polynomial (trinomial) in the form of ax² + bx + c.  We'll start with one demo just using letters, where x is the variable and the other letters are constants:

Ax² + Bx + C

Now, here is what you do for the AC method:

1.  The first step is to multiply the A and the C constants.  Just take the two numbers (see the example that follows), and multiply them.  Keep this value in mind.

2.  Next, we want to find two values whose sum equals B, and which multiply to the product we found for AC.  This might take some trial and error, but when you find these two numbers, let's call the smaller one M and the bigger one N.

3.  Rewrite the original equation (Ax² + Bx + C) by substituting in our values for M and N.  That is, we substitute in Mx + Nx = Bx.  So then, we now have:

Ax² + Mx + Nx + C

4.  Once we have the expression in this form, we can now try to factor by grouping.  (Refer to my earlier post about methods of factoring to review grouping, if you need.  I will also be posting more about this trick soon.)

5.  Rearrange the grouped expression by the distributive property to arrive (hopefully) at your final solution!

This method of factoring will come in quite handy when you come across trinomials, especially ones that have a value for A and don't appear to have an immediately obvious way to factor or simplify.  This AC method of factoring will come in very handy in these situations!

Let's try factoring an example now with real numbers.  Follow along with the steps I've listed above.

Factor:  2x² + 9x + 9
Step 1:  AC = (2)(9) = 18
Step 2:  Two numbers to sum to 9 and multiply to 18.  Pairs can be 1 and 8, 2 and 7, 3 and 6, etc... this should be evident that the pair we like is 3 and 6.  They sum to 9 and multiply to 18.  So, M=3 and N=6
Step 3:  = 2x² + 3x + 6x + 9
Step 4:  = (2x² + 3x) + (6x + 9).  This can be factored to x(2x + 3) + 3(2x + 3).
Step 5:  Rearrange: (2x + 3)(x + 3).

And this would be your solution.  You just used the AC method to factor a trinomial!  I hope that demonstrates this method for you and that you can understand the steps you need to perform.  Let me know if you'd like more examples, and I can put some up!  As always, please +1 me below if this helped you!

Special Polynomial Products

Following up on my previous post about Polynomials in Mathematics, I wanted to dedicate a brief post to special polynomial products.  These are products that will occur very frequently throughout your mathematics studies, and so it is a very good idea to remember these products.

If you would like to refresh on multiplying together polynomials in general, follow the link to view my previous post about polynomials.  Alternately, if you would like to see how to FOIL polynomials, a special method you can use to multiply together binomials, I refer you to my previous post on this as well.

Now, on to the Special Polynomial Products.

These are not overly complicated, and can be derived using basic knowledge of multiplying together polynomials.  But, as I said, these will occur so frequently, that you might as well memorize them so you can save time and not have to worry about deriving them every time you see them.

(A - B)(A + B) = A² - B²

This first one is the easiest to arrive at.  Notice that the signs are OPPOSITES!  Simply apply the FOIL method to the starting expression, and you will arrive at the simplified product.  I'll leave that for you to check for yourself.

(A + B)² = A² + 2AB + B²
(A - B)² = A² - 2AB + B²

This set is also quite simple to see, once you FOIL them out.  The starting terms are squared, you can easily rewrite them as (A+B)(A+B), etc, apply the FOIL method, and come up with the product above.

(A + B)³ = A³ + 3A²B + 3AB² + B³
(A - B)³ = A³ - 3A²B + 3AB² - B³

This set, obviously, is a bit tougher to arrive at, but by tougher I only mean 'requires more work'.  There aren't any tricks and they aren't any more difficult than the others.  I'll derive one for you at the end, after I show you the final set of special products, which is...

(A + B)(A² - AB + B²) = A³ + B³

(A - B)(A² + AB + B²) = A³ - B³

This set requires the most work to derive, so you can see what I mean when I say it pays to memorize these so that you don't have to mess around every time you see them!  I won't derive these ones, but you should try to on your own so that you see how they work and how to arrive at the products IN CASE YOU FORGET THEM!  It's ALWAYS a good idea to understand how to do all these things so that you can do them WITHOUT the shortcuts.

Anyways, now I will get you started by showing you how to multiply (A + B)³ and arrive at the special product I showed you above:

(A+B)³.....rewrite this
(A+B)[(A+B)(A+B)]..... note I've isolated 2 binomials so that I can show you to FOIL easier
(A+B)(A² + 2AB + B²)..... now, multiply the two polynomials together.  Multiply everything in the second term by A, then multiply everything in the second term by B.
(A)(A²) + (A)(2AB) + (A)(B²) + (B)(A²) + (B)(2AB) + (B)(B²)
A³ + 2A²B + AB² + A²B + 2AB² + B³... Now you can combine like terms.
A³ + 3A²B + 3AB² + B³

And there you have it.  It took a little bit of work, but it's not that hard.  You just have to keep going at it.

So, those are the special products of polynomials.  Try to memorize them to make you homework easier, but MORE IMPORTANTLY, please try to work through the derivations and understand HOW to arrive at the products.

How to FOIL Polynomials

This post is just a quick refresher of one of my early posts on polynomials that explains how to FOIL polynomials.  This is an important lesson, so I just want to reiterate the process once more.

You would use the FOIL method when you are trying to multiply two binomials together...that is, an expression in the form (x + A)(x + B).

The term FOIL stands for First Outer Inner Last (sometimes also using Outside and Inside instead).  This refers to the order of multiplying terms together.

I will explain it through an example, which should hopefully be all that is necessary.  Let's try to find the product of (x+4)(x+6).

To apply the FOIL method, you have to first identify and understand what terms we will be talking about.

Here is what the letters in FOIL mean:

The term First means "the first terms in each binomial."  In this case, x and x are the First terms.
The term Outer means "the outer terms of the product expression."  In this case, the first x and 6 are the Outer terms.
The term Inner means "the inner terms of the product expression."  In this case, the 4 and the second x are the Inner terms.
The term Last means "the last terms in each binomial."  In this case, 4 and 6 are the Last terms.

So then, now that you have identified each of the components of FOIL, now all you do is MULTIPLY the two terms for each letter of FOIL, and then add them up (there will be 4 products).

For this example, we'd have these products:

F: (x)(x) = x²

O: (x)(6) = 6x

I: (4)(x) = 4x

L: (4)(6) = 24

Now, we just add them together, combining like terms wherever possible (here, the inner and outer terms):

x² + 6x + 4x + 24

x² + 10x + 24

And that's all there is to the FOIL method.  If you'd like more examples or a better explanation, leave me a comment.  But I think that the example should be sufficient for now to demonstrate the concept.

Cool Math Games

Cool math games are a great way for students to learn valuable math skills without it feeling like studying. These games allow the students to learn and reinforce mathematics concepts that they have been taught, but in the context of a fun game. Obviously, everyone likes having FUN more than having to study. But with educational games, it won't even feel like studying, and they will help develop math skills. Math games can be important learning tools for students of all ages, and should be tried by everyone studying math. \instead of search for math lessons online, trying searching for some of these.  Cool math games will make learning math fun!

Math games are especially useful when teaching young kids various math concepts. Younger students may have a tendency to lose focus on new / hard subjects such as math, and so presenting new concepts using traditional methods may not always be the most productive way of doing things. Perhaps a better plan would be to describe the concept as simply as possible, but then instead of assigning repetitive homework problem sets, allow the students to play math games. These fun activities for children will keep their attention much longer than a boring text book. But as opposed to them wasting their time when they need to take a break from regular studying, they will be developing their math skills while playing these educational games. These games shouldn't replace conventional homework, but fun math games are a great way to compliment it.

Here is a sample of the most popular math games on the internet right now (courtesy of Coolmath.com, one of many game websites):

Bloxorz

Boombot

Fraction Splat

Xfactor

Duck

Concentration

(I'd recommend giving their site a visit to see if you can find a game for the subject you are studying.  It's a great website for kids.)

Another benefit of playing cool math games is that these games develop focus and concentration skills in order for the student to win or succeed. Furthermore, they can help develop competition and communication between students as they try to score more points or achieve higher goals in the games. This motivates them to want to do better and so will strive to understand the concepts. Compare these benefits to the process of having to repeatedly write out math problems (boring!), or get frustrated with new concepts, possibly while studying alone and not in a social environment. From this point of view, cool math games provide many benefits over traditional written homework and studying.

Math games are increasingly becoming more popular amongst both students and teachers. While there are still many teachers who focus on the old-school teaching methods and styles that they know and themselves were taught with, many are now seeing the benefits of fun math games on the students' ability to learn new mathematics concepts. In the end, it shouldn't be about HOW the student is taught these skills, but rather it should only be a matter that they ACHIEVE these skills. Cool math games are a great way to assist in this goal.

Zero Exponent and Negative Exponents

Just a short post to follow up on my last one about the properties of exponents.  In that post, I explained to you various properties of exponents that allow you to do arithmetic with them to simplify them into easier to work with expressions.  However, I only used positive exponents in my explanations.  Here, I need to briefly explain two more concepts: zero exponents and negative exponents.

Zero exponents are simple.  They follow one single rule that applies wherever you will see them.  No matter what the base is, if it is as simple as x, or as complicated as (45xyz / (412ab+26c)), if you raise the base to a power of zero, the expression equals 1.

a⁰ = 1

It is a very handy shortcut, so whenever you see it, you can make life a lot easier for yourself if you just simplify the expression to 1.  For example, you don't want to get stuck rearranging to solve for a in that 45xyz... example above, when you can simplify the whole thing to 1.

So that is zero exponents.

Next is negative exponents.  They aren't QUITE as simple, but the trick to working with them is.  And here it is:

Whenever you see a negative exponent, all you do is make it positive, but put the whole thing on the bottom of a fraction under 1, like so:

a^-n = 1 / aⁿ

Once it's like that, you can apply any or all of the properties of exponents to it without having to deal with the negative anymore.

Just to put some numbers to it, here is an example for you to see what I mean:

(a²b³)^-4
= 1 / (a²b³)⁴
= 1 / a⁸b¹²

Hopefully you can see what I mean with these examples.  They are just a few shortcuts that will make working with exponents a lot easier for you.  Let me know if you need any more explanation about these topics, but for now I will leave it at that.