cookiimonster627
  • cookiimonster627
Expand the binomials using the Binomial theorem. (2x+3y)^4
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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zepdrix
  • zepdrix
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:
zepdrix
  • zepdrix
\[\large\rm =\left(\begin{matrix}n \\ 0\end{matrix}\right)(\color{orangered}{a})^{n}(\color{royalblue}{b})^{0}+\left(\begin{matrix}n \\ 1\end{matrix}\right)(\color{orangered}{a})^{n-1}(\color{royalblue}{b})^{1}+...+\left(\begin{matrix}n \\ n\end{matrix}\right)(\color{orangered}{a})^{0}(\color{royalblue}{b})^{n}\]
zepdrix
  • zepdrix
Looks fancy and confusing I know. But there are some nice patterns going on that we can take advantage of.

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zepdrix
  • zepdrix
The power on the first term is counting DOWN, while the power on the other one is counting UP.
zepdrix
  • zepdrix
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).
cookiimonster627
  • cookiimonster627
Okay
zepdrix
  • zepdrix
\[\large\rm (\color{orangered}{2x}+\color{royalblue}{3y})^4\quad=\sum_{k=0}^4 \left(\begin{matrix}4 \\ \rm k\end{matrix}\right)(\color{orangered}{2x})^{4-k}(\color{royalblue}{3y})^k\]
zepdrix
  • zepdrix
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.
cookiimonster627
  • cookiimonster627
okay
zepdrix
  • zepdrix
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.\[\rm =\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}+\text{___}(\color{orangered}{2x})^{}(\color{royalblue}{3y})^{}\]
zepdrix
  • zepdrix
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?
cookiimonster627
  • cookiimonster627
multiply them @zepdrix
zepdrix
  • zepdrix
Well for the blue stuff, we have the powers `counting up` instead of down.
zepdrix
  • zepdrix
\[\rm =\text{__}(\color{orangered}{2x})^{4}(\color{royalblue}{3y})^{0}+\text{__}(\color{orangered}{2x})^{3}(\color{royalblue}{3y})^{1}+\text{__}(\color{orangered}{2x})^{2}(\color{royalblue}{3y})^{2}+\text{__}(\color{orangered}{2x})^{1}(\color{royalblue}{3y})^{3}+\text{__}(\color{orangered}{2x})^{0}(\color{royalblue}{3y})^{4}\]
zepdrix
  • zepdrix
@cookiimonster627 and then your coefficients in front of each number come from Pascal's Triangle.

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