Question

Expand the following using binomial theorem: (2x+4)^3

I don't know how to do the binomial theorem.

Answer 1

\[(a+b)^{n}=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k}b ^{k}\]\[\left(\begin{matrix}n \\ k\end{matrix}\right)\] is read as 'n over k' and is calculated as
\[\frac{ n! }{ k!*(n-k!) }\]
To summarize, the solution to your problem is
\[8x ^{3}+6*4x ^{2}*4+6*2x*16+64 => 8x ^{3}+64x ^{2}+192x+64\]

Answer 2

To expand on that a bit (and correct an error):

you have \[a = 2x\]\[b=4\]
Now you compute your binomial coefficients from 0 to 3:

\[\frac{3!}{0!(3-0)!} = 1\]\[\frac{3!}{1!(3-1)!} = \frac{3*2*1}{1*2*1} = 3\]\[\frac{3!}{2!(3-2)!} = \frac{3*2*1}{2*1*1} = 3\]\[\frac{3!}{3!(3-3)!} = \frac{3*2*1}{3*2*1*1} = 1\]

Now fill in the blanks:
\[(2x+4)^3 = (a+b)^3 = 1*a^3b^0 + 3*a^2b^1+3*a^1b^2 + 1*a^0b^3\]\[=1*(2x)^3(4)^0+3*(2x)^2(4)^1+3*(2x)^1(4)^2+1*(2x)^0(4)^3\]\[=8x^3+48x^2+96x+64\]