Question

Expand the binomials using the Binomial theorem.
(2x+3y)^4

Answer 1

Hey cookie :)

Here is our Binomial Theorem:\[\large\rm (\color{orangered}{a}+\color{royalblue}{b})^n\quad=\sum_{k=0}^n \left(\begin{matrix}n \\ \rm k\end{matrix}\right)\color{orangered}{a}^{n-k}\color{royalblue}{b}^k\]Which expands out like this:

Answer 2

\[\large\rm =\left(\begin{matrix}n \\ 0\end{matrix}\right)(\color{orangered}{a})^{n}(\color{royalblue}{b})^{0}+\left(\begin{matrix}n \\ 1\end{matrix}\right)(\color{orangered}{a})^{n-1}(\color{royalblue}{b})^{1}+...+\left(\begin{matrix}n \\ n\end{matrix}\right)(\color{orangered}{a})^{0}(\color{royalblue}{b})^{n}\]

Answer 3

Looks fancy and confusing I know.
But there are some nice patterns going on that we can take advantage of.

Answer 4

The power on the first term is counting DOWN,
while the power on the other one is counting UP.

Answer 5

We have a 4th power,
If we're counting from 0 to 4, that's a total of 5 terms,
(because we include the 0 term).

Answer 6

Okay

Answer 7

\[\large\rm (\color{orangered}{2x}+\color{royalblue}{3y})^4\quad=\sum_{k=0}^4 \left(\begin{matrix}4 \\ \rm k\end{matrix}\right)(\color{orangered}{2x})^{4-k}(\color{royalblue}{3y})^k\]

Answer 8

So we'll expand this out,
here is where things get a little tricky.
You need to apply the power which are counting down to `all of the orange` part.
Not just the x.

Answer 9

okay

Answer 10

Let's not use the weird n choose k thing in front,
let's just use a space for now.
Here is a nice way to initially set up your problem.\[\rm =\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}\]

Answer 11

So like I was saying, our powers of the orange should count DOWN from the full value,\[\rm =\text{___}(\color{orangered}{2x})^{4}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{3}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{2}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{1}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{0}(\color{royalblue}{3y})^{}\]ok with that step? :)

What will we do with the blue parts?
Any ideas?

Answer 12

multiply them @zepdrix

Answer 13

Well for the blue stuff, we have the powers `counting up` instead of down.

Answer 14

\[\rm =\text{__}(\color{orangered}{2x})^{4}(\color{royalblue}{3y})^{0}+\text{__}(\color{orangered}{2x})^{3}(\color{royalblue}{3y})^{1}+\text{__}(\color{orangered}{2x})^{2}(\color{royalblue}{3y})^{2}+\text{__}(\color{orangered}{2x})^{1}(\color{royalblue}{3y})^{3}+\text{__}(\color{orangered}{2x})^{0}(\color{royalblue}{3y})^{4}\]

Answer 15

@cookiimonster627 and then your coefficients in front of each number come from Pascal's Triangle.