CLEP College Algebra/Binomial Theorem

Quick question: what is $(2x-4{)}^2$ equivalent to? Well, to find that, we need to multiply $(2x-4)$ two times because an exponent denotes how many times, $n$, you multiply any value $a$, which we represent as $a^n$. Let $a=(2x-4)$ and $n=2$.

$a^n=(2x-4{)}^2=(2x-4)(2x-4)=4x^2+-8x+16=4(x^2-2x+4)$.

If this was part of a multiple choice question, you would perhaps be able to quickly get the answer. However, what if you had to do a modified version of this problem?

Find $(2x-4{)}^{10}$.

You can perhaps do this problem after a while, but it might already be too late before you can finish the CLEP College Algebra exam. Even if it is not too late, you may not be able to do the other problems after you are done, or you may get the wrong answer because it is very hard to do the problem like that. Therefore, how do we do the problem? That is when the binomial expansion theorem comes into play. We will state the definition once we understand the foundations of the formula first.

Permutations and Combinatorics

Factorial Notation

Let's start small. Try to imagine a list of all the possible $1$-letter words using the standard alphabet. In that list, you would see all the letters of the alphabet.

Now try to imagine a list of all the possible $2$-letter words you can make using the standard alphabet. If you think about it, that is simply $26\text{ letters}\cdot 25\text{ letters}$. After all, you had $26$ letters to choose from, but since you are not allowed to use that letter more than once, you now have $25$ letters to choose from. This works the same for any combination of $2$-letter words (AB, AC, AD, $\dots$, BA, BC, $\dots$, ZA, ZB, $\dots$, ZY).

If you think about it, the pattern continues. The number gets larger, but it gets larger by multiplying the number from before minus $1$. If you want to make a random word that is $26$ letters long, where no letters can be repeated more than once, then
$$\underset{26}{\underbrace{26\cdot 25\cdot 24\cdots 2\cdot 1}}\text{ words can be created from the standard English alphabet.}$$Template:Robox/Close

In Example 2.1, we wanted to find the number of 26-letter words you can make, if you can only use a letter once. It can be a bit tedious to write $\underset{26}{\underbrace{26\cdot 25\cdot 24\cdots 2\cdot 1}}$ as your answer. However, mathematicians have found a way to write a short hand for it.

To make the above definition more clear, here is a list of a few examples:

$$\begin{array}{cc}x!& f(x)\\ 2!& \underset{2}{\underbrace{2\cdot 1}}=2\\ 3!& \underset{3}{\underbrace{3\cdot 2\cdot 1}}=6\\ 4!& \underset{4}{\underbrace{4\cdot 3\cdot 2\cdot 1}}=24\\ 5!& \underset{5}{\underbrace{5\cdot 4\cdot 3\cdot 2\cdot 1}}=120\\ ⋮& ⋮\end{array}$$

In Example 1.1, the answer can be rewritten as this:
$$\underset{26}{\underbrace{26\cdot 25\cdot 24\cdots 2\cdot 1}}=26!$$

We use the factorial notation much more than just for a representation of multiplication in specific situations. The factorial is very useful when listing out possibilities or the number of arrangements. Say that you want to find the number of ways you can arrange 4 students in a group. Call each student either A, B, C, or D, only once. Here are some listings:

• A, B, D, C
• A, B, C, D
• A, C, B, D
• A, C, D, B
• A, D, B, C
• A, D, C, B

As we continue listing each arrangement, we find that the number of ways in which you can arrange the four students is $24$, which is equal to $4!$. This makes sense because you cannot have an arrangement in which a student repeats in the list (no cloning allowed!).

Say that you want to find the number of ways you can arrange three students in a group. Call each student either B, C, or D, only once. Here are all of the listings of students.

• B, D, C
• B, C, D
• C, B, D
• C, D, B
• D, B, C
• D, C, B

The number of ways in which you can arrange the three students is $6$, which is equal to $3!$.

From the two examples above, it should be clear that factorial notation is very useful because we can show that arrangements are simply just a choice between $n$ possibilities, then $(n-1)$ possiblities, etcetera.

Combination

Let us imagine a group of 15 teachers. Each one will be assigned a letter from the English Alphabet, in order from A - O. Instead of adding 6 teachers to the committee, let us add 3 teachers, so that we get "a feel" for the situation.

In a committee, we do not care about the order in which we find them at a table because we only care about if they are in the committee. Imagine we see three teachers, A, B, and C.

• ABC
• ACB
• CAB
• CBA
• BAC
• BCA

The list you see above does not tell us that there are different teachers, only that the order is different. However, we do not care about the order; therefore, we must exclude the order from the number of possible committees to make the answer representative of the idea of a committee. The numbers of arrangements in which the same number of teachers, $x$, show is $x!$. This is what we "factor out," the order in which we find teachers in committees.

One final thing to keep in mind is that since we are selecting $6\text{ teachers}$ from a group of $15\text{ teachers}$, we are excluding $15-6\text{ teachers}$ or $9\text{ teachers}$ from being arranged into the committee in the same way we expect them to. Therefore, we must "factor out" the $(15-6)!$ ways of arranging those teachers. From this, we learn that we can arrange
$$\frac{15!}{(15-6)!6!}=\frac{15!}{9!\cdot 6!}=5005\text{ teachers into the new committee.}$$Template:Robox/Close

One difference between Example 1.1 and Example 1.2 is that the order in Example 1.2 did not matter. After all, in whichever way you arrange the committee, you count any arrangement where the same elements are within identical to the other. You cannot do that with an arrangement of letters because AB and BA are not identical. Nevertheless, this example is not independent. There are countless times in which you whatever way you arrange a group of something, the number of possibilities does not increase. Therefore, we need a formula and notation for such a situation.

We are now closer one step closer to understand the binomial theorem. We only need to understand a few more things before we use the formula we learned at the very beginning.

Polynomials and Binomials

Polynomials and Generalization

Look at the following polynomials below:

$2x^2+5x-3$

$-3x^5+4x^4+5x^3-6x$

$\frac{12}{19}{x}^3+\frac{5}{3}{x}^2+\frac{1}{12}x-\frac{3}{2}$

The one that is in common with all three of the polynomials above is that they are all added (remember, the inverse of summation for real number $x$ is $-x$, so $x+(-x)=0$). This means that we have a series. To really hammer that idea into our skulls, we need to go ahead and generalize a polynomial. Here is a formal definition of a polynomial

Although the definition may seem complicated, there is no need to worry. Remember that every polynomial must have a degree, in this instance, $i$. For every $x$, the coefficient is $a_i$. If $i=0$, then $a_i{x}^i=a_0{x}^0=1\cdot a_0=a_0$. Since each iteration of $i$ has a different coefficient, once $i=k$, we are finished writing the series. Every polynomial, therefore, has the same pattern.

However, while this may be an incredible mathematical fact, this does not help us with our original problem. After all, we want to be able to predict a polynomial that results from multiplying another polynomial (since a binomial is also a polynomial). However, the above definition gives us a giant hint to help us solve our first problem. Before we create a formula for that situation, however, we must understand one more concept.

Pascal's Triangle

French mathematician Blaise Pascal, like many mathematicians before his time, wanted to study a pattern of sequences and series. However, unlike most patterns we see in this WikiBook, the pattern is not horizontal, but vertical and diagnol. Instead of one-dimensional patterns, we see a two dimensional pattern. This pattern is of a shape of a triangle, which is shown below.