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Algebra II: High School23 chapters | 203 lessons

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Lesson Transcript

Instructor:
*Maria Airth*

Maria has a Doctorate of Education and over 15 years of experience teaching psychology and math related courses at the university level.

In this lesson, students will learn the binomial theorem and get practice using the theorem to expand binomial expressions. The theorem is broken down into its parts and then reconstructed.

Welcome to this lesson on the ** binomial theorem**. While the term sounds complicated, breaking it down into its parts can make it easily understandable:

Since we know that a **binomial** is a 2-term expression, and a **theorem** is a mathematical formula, **binomial theorem** must mean a mathematical formula used to expand 2-term expressions. It is used to find the expanded version of binomials raised to any numerical exponent.

The theorem looks like this:

Before I move on, I'd like to cover three more definitions that will be important:

**Term**is the combination of variables and coefficients separated by operations like + or -**Variable**is the unknown of a term, usually represented by letters**Coefficient**is the number being multiplied by the variables

In 3*ab* + 2*b* there are two terms, 3*ab* and 2*b*; two variables, *a* and *b*; and two coefficients, 3 and 2.

Back to the formula. Don't let it intimidate you! Even though it seems overly complicated and not worth the effort, the binomial theorem really does simplify the process of expanding binomial exponents. Just think of how complicated it would be to expand 55 manually. You would have to multiply each step on your own:

5 x 5 = 25

5 x 25 = 125

5 x 125 = 625

5 x 625 = 3125

And that is only to the fifth power! Can you imagine how long it takes if you want to go further? I'm certainly glad that calculators have the ability to do this calculation in a fraction of a second. That means I don't have to.

But, if a single number can get so bogged down in the process to expand manually, what if it were a binomial instead? What if we had (*a* + *b*)2. We would have to calculate (*a* + *b*) (*a* + *b*). Well, that isn't so bad; if you are familiar with squaring binomials, you will quickly come to the answer of *a*2 + 2*ab* + *b*2. (If you aren't sure how I got that answer, please review the lessons covering multiplying terms.)

While squaring a binomial isn't terribly difficult or time consuming, what about (*a* + *b*)5. Here you would have to do the same thing as we did with the 5: multiply each answer times the original binomial until you have multiplied the binomial by itself five times.

We already have the first two, giving us a result of *a*2 + 2*ab* + *b*2. Now to get (*a* + *b*) (*a*2 + 2*ab* + *b*2), we have to remember that we must multiply each term in the binomial by each term in the trinomial (that is the bracket with three terms in it). Let's start with the *a* term: *a* * (*a*2 + 2*ab* + *b*2 ) = *a*3 + 2*a*2 *b* + *ab*2.

Then the *b* term: *b* * (*a*2 + 2*ab* + *b*2 ) = *a*2 *b* + 2*ab*2 + *b*3.

Then we have to combine like terms to get a final result of *a*3 + 3*a*2 *b* + 3*ab*2 + *b*3

And that gives us the result for (*a* + *b*)3; just two more multiplyings to go, and we are done with raising the binomial to the fifth power! I don't know about you, but I'm already tired of doing this manual process for expanding a binomial exponent, and we've only just finished the third power.

That is what the theorem is good for. It allows us to work out a formula to get straight to our end result instead of having to continually repeat the binomial multiplication.

Let's take a closer look at the formula now.

Starting at the beginning, we have (*a* + *b*)*n*. This just lets us know what power has been chosen for the binomial. *n* will be defined in each real problem. To **define** a variable means to assign an actual value to a variable. For our process, let's define *n* as 3. The equals sign lets us know that the formula is coming next.

Now, that funny angular *E* is called **sigma** and means that we will have to repeat a process and sum all the results. Below the sigma, you see *k* = 0; this just indicates that we will start our summation with the understanding that *n* can be any number from 0 to whatever it has been defined as. Above the sigma, we see *n* again. This time, it indicates that the highest exponent is to be used.

Next is what looks like an overgrown parenthesis with a fraction inside. But, the fraction is missing its fraction bar. This is actually the mathematical notation for *'n* choose *k*' and means that you create a fraction out of the factorials of *n* and *k*, like this: *n*! / *k*!(*n* - *k*)! The factorial fraction will give us the coefficient for each term in the final expanded result.

Finally, the remainder of the formula will result in the correct exponents attached to the variables for each term. Notice that the *a* variable will start at the max value and work its way to zero, while the *b* variable does the opposite.

Also remember:

- Anything0 = 1
- 0! = 1

And that is the formula in a nutshell:

- Introducing the binomial and its exponent
- Indicating that multiple results of a process will be summed
- The coefficient of each term is defined by
*'n*choose*k*' - The exponents for each term are determined by a pattern of reducing the first by 1 and increasing the second by 1 consecutively

If we use the example here, of (*a* + *b*)3, our coefficient calculations would be:

- For the first term, the factorial calculation would be 3! / 0! (3 - 0)!, which works out to 1.
- For the second term, the factorial calculation would be 3! / 1! (3 - 1)!, which works out to 3.
- For the third term, the factorial calculation would be 3! / 2! (3 - 2)!, which is 3.
- Finally, for the fourth term, the factorial calculation would be 3! / 3! (3 - 3)!, which is 1.

So, our coefficients for a binomial raised to the third power are 1, 3, 3, 1.

If you aren't quite sure how I got the factorial calculations, please review other lessons.

Now that we know all the coefficients for our terms, we just need to figure out the exponents for our variables. The formula says that we start with the first variable having the greatest exponent value and the second variable starting at 0 (which means that it will not appear in the term).

Again working with our example, we would get:

- for the
*first term*,*a(3 - 0) b0*, which is just*a3* - the
*second term*would be,*a(3 - 1) b1*, which looks like*a2 b*when simplified - the
*third term*would be,*a(3 - 2) b2*, which is*ab2* - and last is the
*fourth term*,*a(3 - 3) b3*, or simply*b3*

So, our variables are: *a3 a2 b ab2 b3*

__video editor note: please fill in the variables from above into the line segments at the top of the screen and increase the size of the final expansion to fill the screen.__

Putting together the coefficients and variables with exponents from our example we have:

*(a + b)3 = a3 + 3a2 b + 3ab2 + b3*

Let's work through one more example. This time, instead of separating the calculations by part, I will do all the calculations for each term as it comes.

Our example this time is *(x + y)4 * so we start with the first term:

(*4! / 0!(4 - 0)!) x4 *, which is really just *x4 *

Next we have**4! / 1!(4 - 1)!) x3y*, which works out to *4x3y*

Then, *(4! / 2!(4 - 2)!) x2 y2* or *6x2 y2*

Next is *(4! / 3!(4 - 3)!) xy3* or *4xy3*

And finally we have*(4! / 4! (4 - 4)!) x0 y4*, which is *1y4*

So, *(x + y)4 = x4 + 4x3 y + 6x2 y2 + 4xy3 + y4*.

So, there you have it - it may seem like it takes a long time to get to the answer, but as you practice your factorials and get into that easy pattern of subtracting one from the first variable and adding one to the second on each term, it will become really easy to expand binomials with any exponent.

In this lesson, we learned that a **binomial theorem** is just a formula for expanding two terms raised to any exponent. While the formula looks a bit complicated, it can be divided into its parts to understand it better:

The first section is simply the binomial with the exponent defined. Next, you find sigma indicating to sum the results of the formula for values from 0 to the defined exponent. Third, you find the notation for *'n* choose *k*,' which indicates using factorial calculations to get coefficients. And, finally, the last section of the formula calculates the exponents attached to each variable in each term and can be found by simply adding one and subtracting one from each variable in each term respectively.

Don't forget, practice makes perfect and while this formula seems complicated on the face of it, each individual part requires very simple math skills that are all put together at the end to create a much easier way to expand a binomial than using labor-intensive manual multiplying.

Using the information gained from this video lesson, you could:

- Identify the binomial theorem
- Understand the advantage of using this theorem
- Calculate the binomial theorem formula
- Find coefficients

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Algebra II: High School23 chapters | 203 lessons

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