# How do you use the Binomial Theorem to expand (x + 2)^4?

Jan 27, 2017

${\left(x + 2\right)}^{4} \text{ "=" } {x}^{4} + 8 {x}^{3} + 24 {x}^{2} + 32 x + 16$.

#### Explanation:

The Binomial Theorem says that, for a positive integer $n$,

${\left(x + b\right)}^{n} = {\text{_nC_0 x^n + ""_nC_1 x^(n-1)b+""_nC_2 x^(n-2)b^2 + ... + }}_{n} {C}_{n} {b}^{n}$

This can be succinctly written as the sum

(x+b)^n= sum_(k=0)^n ""_nC_k x^(n-k)b^k

To see why this formula works, let's use it on the binomial for this question, ${\left(x + 2\right)}^{4}$. If we were to write out all the factors side-by-side, we'd get

${\left(x + 2\right)}^{4} = \left(x + 2\right) \left(x + 2\right) \left(x + 2\right) \left(x + 2\right)$

Multiplying this out means making all possible products of one term from each binomial, and adding these products together. Sort of like FOIL-ing to the next level. (In FOIL-ing, there are $\textcolor{b l u e}{2}$ binomials, so there will be ${2}^{\textcolor{b l u e}{2}} = 4$ terms; with $\textcolor{g r e e n}{4}$ binomials, there will be ${2}^{\textcolor{g r e e n}{4}} = 16$ terms.)

After (tediously) multiplying it all out, we'll get some terms with ${x}^{4}$, some with ${x}^{3} \cdot 2$, some with ${x}^{2} \cdot {2}^{2}$, and so on. But how many of each?

Since there is only ""_4C_0=1 way to pair up all four $x$'s, we'll only get a single ${x}^{4}$ term.

There are ""_4C_1=4 ways to make an ${x}^{3} \cdot 2$ term—using the $2$ from binomials 1, 2, 3, and 4 (multiplied with the $x$'s from the others).

Continuing this way, there will always be ""_4C_k ways to make an ${x}^{n - k} \cdot {2}^{k}$ term. So this is where the combination notation in the Binomial Theorem comes from. For $k = 0 , \ldots , 4$, the ${x}^{n - k} \cdot {2}^{k}$ bit (from the sum above) represents the different terms we can make, and the ""_4C_k bit is the number of ways to make that term.

Thus, the binomial expansion of ${\left(x + 2\right)}^{4}$ is

${\text{_4C_0 x^4 + ""_4C_1 x^3 * 2 + ""_4C_2 x^2 * 2^2 + ""_4C_3 x * 2^3 + }}_{4} {C}_{4} \cdot {2}^{4}$

$= 1 {x}^{4} + 4 {x}^{3} \left(2\right) + 6 {x}^{2} \left(4\right) + 4 x \left(8\right) + 1 \left(16\right)$

$= {x}^{4} + 8 {x}^{3} + 24 {x}^{2} + 32 x + 16$.