A binomial is a polynomial with two terms. We're going to look at the Binomial Expansion Theorem, a shortcut method of raising a binomial to a power.

(x+y)^{0} = 1

(x+y)^{1} = x + y

(x+y)^{2} = x^{2} + 2xy + y^{2}

(x+y)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + y^{3}

(x+y)^{4} = x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}

(x+y)^{5} = x^{5} + 5x^{4}y + 10x^{3}y^{2} +10x^{2}y^{3} + 5xy^{4} + y^{5}

There are several things that you hopefully have noticed after looking at the expansion

- There are n+1 terms in the expansion of (x+y)
^{n} - The degree of each term is n
- The powers on x begin with n and decrease to 0
- The powers on y begin with 0 and increase to n
- The coefficients are symmetric

Pascal's Triangle, named after the French mathematician Blaise Pascal is an easy way to find the coefficients of the expansion.

Each row in the triangle begins and ends with 1. Each element in the triangle is the sum of the two elements immediately above it.

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1

Combinations will be discussed more fully in section 7.6, but here is a brief summary to get you going with the Binomial Expansion Theorem.

A combination is an arrangement of objects, without repetition, and order not being important. Another definition of combination is the number of such arrangements that are possible.

The n and r in the formula stand for the total number of objects to choose from and the number of objects in the arrangement, respectively.

The quickest way, if unimaginative, is to use the combination function on
your calculator. On the TI-82 and TI-83, It is found under the Math menu, Probability
submenu, choice 3. Enter the value for n first, then the
_{n}C_{r} notation, then the value for r.

Each element in Pascal's Triangle is a combination of n things. The value for r begins with zero and works its way up to n. Or, because of symmetry, you could say it begins with n and works its way down to 0.

Let's consider the n=4 row of the triangle.

_{4}C_{0} = 1, 4C_{1} = 4, 4C_{2} = 6, 4C_{3} = 4, 4C_{4} = 1

Notice that the 3^{rd} term is the term with the r=2. That is, we begin counting with 0. This will
come into play later.

Okay, now we're ready to put it all together.

The Binomial Expansion Theorem can be written in summation notation, where it is very compact and manageable.

Remember that since the lower limit of the summation begins with 0, the 7^{th} term of the sequence
is actually the term when k=6.

When you go to use the binomial expansion theorem, it's actually easier to
put the guidelines from the top of this page into practice. The x starts off
to the n^{th} power and goes down by one each time, the y starts off to the 0^{th} power (not there) and increases by one each time. The coefficients are combinations.

Expand ( 3x - 2y )^{5}

Start off by figuring out the coefficients. Remember that these are combinations of 5 things, k at a time, where k is either the power on the x or the power on the y (combinations are symmetric, so it doesn't matter).

C(5,0) = 1; C(5,1) = 5; C(5,2) = 10; C(5,3) = 10; C(5,4) = 5; C(5,5) = 1

Now throw in the 3x and -2y terms.

1(3x)^{5}(-2y)^{0} + 5(3x)^{4}(-2y)^{1} + 10(3x)^{3}(-2y)^{2} + 10(3x)^{2}(-2y)^{3} + 5(3x)^{1}(-2y)^{4} + 1(3x)^{0}(-2y)^{5}

Raise the individual factors to their proper powers.

1(243x^{5})(1) + 5(81x^{4})(-2y) + 10(27x^{3})(4y^{2}) + 10(9x^{2})(-8y^{3}) + 5(3x)(16y^{4}) +
1(1)(-32y^{5})

Simplify each term to get the final answer.

243x^{5} - 810x^{4}y + 1080x^{3}y^{2} - 720x^{2}y^{3} + 240xy^{4} - 32y^{5}

Find the 9th term in the expansion of (x-2y)^{13}

Since we start counting with 0, the 9th term is actually going to be when k=8. That is, the power on the x will 13-8=5 and the power on the -2y will be 8. The coefficient is either C(13,8) or C(13,5), combinations are symmetric, so it doesn't matter.

C(13,8) * (x)^{5} (-2y)^{8 }= 1287 (x^{5}) (256y^{8})
= 329472x^{5}y^{8}