Binomial Theorem

The Equation

The Binomial Theorem is one way to expand binomials without FOIL-ing. It works best when the binomial is raised to a higher degree, but the basic formula is:

$(a + b)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) a^{n - k} b^{k}$

Where $(\binom{n}{k})$ is defined as

$\frac{n!}{k! (n - k)!}$

Ok, so that looks like a lot! Let’s take it section by section.

The Breakdown

The binomial is $(a + b)^{n}$ . That’s a standard binomial.

The standard binomial would usually be expanded as $a^{n} + 2 a b + b^{n}$ , but what if it was raised to the 10th, 20th, or 50th power?! That would be a lot of FOIL-ing, and it would get really hard to keep track of all the numbers! With the binomial theorem, we expand a little differently. The first half of the expansion is the sum of $n$ choose $k$ , for some $n$ , with values of $k$ starting at zero. That looks like this:

$\sum_{k = 0}^{n} (\binom{n}{k})$

The numbers $(\binom{n}{k})$ are also called binomial coefficients, and we will see why later in this post. The second half of the expansion is $a^{n - k} b^{k}$ .

We’re going to keep the theorem in this broken up state for the post, because I find it easier to work with that way. You can use the whole theorem at once if you want, this is just what I do.

Solving Problems

2nd degree binomial

We are going to start with a standard binomial, so you can see the theorem in action. We’ll also expand using FOIL, just to double check that it’s correct.

$(a + b)^{2}$

In this problem, $n$ is equal to $2$ . We will get three different binomial coefficient numbers ( $(\binom{n}{k})$ ). Looking at our formula for $n$ choose $k$ , we know that these three are: $(\binom{2}{0}) = \frac{2!}{0! (2 - 0)!}$

$(\binom{2}{1}) = \frac{2!}{1! (2 - 1)!}$ $(\binom{2}{2}) = \frac{2!}{2! (2 - 2)!}$ So now, we solve the factorials to get our coefficient. The first one is equal to $1$ , the second one is equal to $2$ , and the third is equal to $1$ .

0!

$0!$ is equal to $1$ , because there is exactly one way to order a group of zero numbers.

That’s the first part of the theorem, now we need to apply it to the second part: $a^{n - k} b^{k}$ We know the value of $n$ , and we know the values of $k$ ( $0$ , $1$ , and $2$ ), so now we just need to plug them in. For $k = 0$ , this part of the equation will only be equal to $a^{2}$ . We are subtracting nothing from the $n$ , and $b$ is raised to no powers. When $k = 1$ , it will be equal to $a b$ , and when $k = 2$ , it is only equal to $b$ . However, our final expansion is NOT $a^{2} + a b + b^{2}$ . We are missing the first half of the equation, and we need to apply our binomial coefficient numbers we found. When we apply the coefficients, it looks like this:

$\begin{aligned} (\binom{2}{0}) & = 1 \\ a^{2 - 0} b^{0} & = a^{2} \\ (\binom{2}{1}) & = 2 \\ a^{2 - 1} b^{1} & = 2 a b \\ (\binom{2}{2}) & = 1 \\ a^{2 - 2} b^{2} = a^{0} b^{2} & = b^{2} \end{aligned}$ This makes the full expansion of the binomial $(a + b)^{2}$ $a^{2} + 2 a b + b^{2}$ The FOIL method produces the same results: $\begin{aligned} (a + b)^{2} & = (a + b) (a + b) \\ a (a + b) & = a^{2} + a b \\ b (a + b) & = b^{2} + a b \\ a^{2} + a b + a b + b^{2} & = a^{2} + 2 a b + b^{2} \end{aligned}$ While both ways seem about the same when raised to the 2nd power, when we started raising more and more, the binomial theorem method will become a very useful tool. Even for powers of 3 or 4, FOIL-ing can get a little bit annoying.

3rd degree binomial

When higher degrees and more coefficients were introduced, I very possibly cried, so I hope to avoid that here!

Let’s tackle the binomial $(3 a - b)^{3}$ I’ll be FOIL-ing at the end of this section, but first, we will use the binomial theorem to do the expansion of this. We have our $n$ , and we know where to start with $k$ , so we can dive right in!

First, we need to do the sum of $(\binom{3}{k})$ with values of $k$ ranging from $0$ to $3$ . That will look like this:

$(\binom{3}{0}) = \frac{3!}{0! (3 - 0)!}$

$(\binom{3}{1}) = \frac{3!}{1! (3 - 1)!}$ $(\binom{3}{2}) = \frac{3!}{2! (3 - 2)!}$ $(\binom{3}{3}) = \frac{3!}{3! (3 - 3)!}$ We know we will have 4 coefficients. The first is $1$ , the second is $3$ , the third is $3$ , and the fourth is $1$ . But we need to remember that we are NOT done. We need to add the second part of the equation, and then re-apply the coefficients (unless you did the entire thing at once).

Our $n$ is $3$ , and we have our values of $k$ ranging from $0$ to $3$ , and all four of our coefficients.

$\begin{aligned} (\binom{3}{0}) & = 1 \\ 3 a^{3 - 0} b^{0} & = 27 a^{3} \\ (\binom{3}{1}) & = 3 \\ 3 a^{3 - 1} * - b^{1} & = - 27 a^{2} b \end{aligned}$ Because $b$ is negative, we have to pay special attention to the signs we use.

$\begin{aligned} (\binom{3}{2}) & = 3 \\ 3 a^{3 - 2} * - b^{2} & = 9 a b^{2} \\ (\binom{3}{3}) & = 1 \\ 3 a^{3 - 3} * - b^{3} & = - b^{3} \end{aligned}$ Fully expanded, it will look like this:

$27 a^{3} - 27 a^{2} b + 9 a b^{2} - b^{3}$ With FOIL, we get the same results.

$(3 a - b)^{3} = (3 a - b) (3 a - b) (3 a - b)$ $\begin{aligned} (3 a - b) (3 a - b) = 9 a^{2} - 6 a b + b^{2} & = \\ (3 a - b) 9 a^{2} - 6 a b + b^{2} & = \\ 27 a^{3} - 18 a^{2} b + 3 a b^{2} - 9 a^{2} b + 6 a b^{2} - b^{3} & = \\ 27 a^{3} - 27 a^{2} b + 9 a b^{2} - b^{3} \end{aligned}$

Now, it may look like FOIL is faster, but this is FOIL typed out, after I did it entirely by hand! Even for binomials raised to the third degree, the binomial theorem makes expansion much faster.

10th degree binomial

This is the last example problem for the more algebraic side of things, and I will not be FOIL-ing at the end for obvious reasons (I’m not about to do that much multiplication).

Let’s take the problem

$(x + y)^{10}$ We have all the values needed for the binomial theorem, so now we will solve for the first half,

$\sum_{k = 0}^{10} (\binom{10}{k})$

All solutions for

\sum_{k = 0}^{10} (\binom{10}{k})

$\begin{aligned} (\binom{10}{0}) & = \frac{10!}{0! (10 - 0)!} = & 1 \\ (\binom{10}{1}) & = \frac{10!}{1! (10 - 1)!} = & 10 \\ (\binom{10}{2}) & = \frac{10!}{2! (10 - 2)!} = & 45 \\ (\binom{10}{3}) & = \frac{10!}{3! (10 - 3)!} = & 120 \\ (\binom{10}{4}) & = \frac{10!}{4! (10 - 4)!} = & 210 \\ (\binom{10}{5}) & = \frac{10!}{5! (10 - 5)!} = & 252 \\ (\binom{10}{6}) & = \frac{10!}{6! (10 - 6)!} = & 210 \\ (\binom{10}{7}) & = \frac{10!}{7! (10 - 7)!} = & 120 \\ (\binom{10}{8}) & = \frac{10!}{8! (10 - 8)!} = & 45 \\ (\binom{10}{9}) & = \frac{10!}{9! (10 - 9)!} = & 10 \\ (\binom{10}{10}) & = \frac{10!}{10! (10 - 10)!} = & 1 \end{aligned}$

Now, we have our coefficients, and we can apply the second half of the theorem.

Second half of theorem with coefficients

(\binom{10}{k})

$\begin{aligned} x^{10 - 0} y^{0} & = x^{10} \\ x^{10 - 1} y^{1} & = 10 x^{9} y \\ x^{10 - 2} y^{2} & = 45 x^{8} y^{2} \\ x^{10 - 3} y^{3} & = 120 x^{7} y^{3} \\ x^{10 - 4} y^{4} & = 210 x^{6} y^{4} \\ x^{10 - 5} y^{5} & = 252 x^{5} y^{5} \\ x^{10 - 6} y^{6} & = 210 x^{4} y^{6} \\ x^{10 - 7} y^{7} & = 120 x^{3} y^{7} \\ x^{10 - 8} y^{8} & = 45 x^{2} y^{8} \\ x^{10 - 9} y^{9} & = 10 x y^{9} \\ x^{10 - 10} y^{1} 0 & = y^{10} \end{aligned}$

SO, fully expanded, its going to be one very large equation, which is why using the binomial theorem is better, so you don’t get tangled up in a bunch of multiplication.

$x^{10} + 10 x^{9} y + 45 x^{8} y^{2} + 120 x^{7} y^{3} + 210 x^{6} y^{4} + 252 x^{5} y^{5} + 210 x^{4} y^{6} + 120 x^{3} y^{7} + 45 x^{2} y^{8} + 10 x y^{9} + y^{10}$

The Binomial Theorem offers a faster way to expand binomials raised to higher degrees, and is a really important theorem to understand.