# How do I use the the binomial theorem to expand (t + w)^4?

Sep 28, 2015

${\left(t + w\right)}^{4} = {t}^{4} + 4 {t}^{3} {w}^{1} + 6 {t}^{2} {w}^{2} + 4 {t}^{1} {w}^{3} + {w}^{4}$

#### Explanation:

First, you should get to know Pascal's triangle. Here's a small portion of it up to 6 lines. Basically you're just adding the numbers that are beside each other, then writing the sum below them. (For example, $1 + 2$ in the third line equals to $3$ in the fourth line.) If you want a better explanation for that, feel free to do some more research on it.

Okay, so take a look at the triangle. The numbers in each line are the coefficients of the terms when a binomial is raised to a certain power. The first line is for ${n}^{0}$, the second is for ${n}^{1}$, the third is for ${n}^{2}$, etc.

For ${\left(t + w\right)}^{4}$, we'll use the 5th line (1 4 6 4 1). Let's write it down:
$1 t w + 4 t w + 6 t w + 4 t w + 1 t w$.

Now, for the first term, the exponent of the first variable $t$ will be $4$ and it will descend in the next terms. The exponent of $w$ will start at $0$ and ascend in the next terms.
$1 {t}^{4} {w}^{0} + 4 {t}^{3} {w}^{1} + 6 {t}^{2} {w}^{2} + 4 {t}^{1} {w}^{3} + 1 {t}^{0} {w}^{4}$

Simplify:
${t}^{4} + 4 {t}^{3} {w}^{1} + 6 {t}^{2} {w}^{2} + 4 {t}^{1} {w}^{3} + {w}^{4}$