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Binomial Theorem will follow this pattern:
The powers start at 3 and decrease down to 0 for the first term,
and start at 0 and increase to 3 for the other term.
(2x+4)^3 =

See how the power counts down on the (2x)
and counts up on the 4?
That's the pattern that our expansion must follow.
I left a little space in front of each term.
The coefficient in front of each term will come from the fourth row of Pascal's Triangle. The one that looks like this:
1 3 3 1
Those are the coefficients we want:

We can clean this up a little bit by getting rid of some of the junk. Anything to the 0th power is 1. So let's suppress all of our 1's because multiplying by 1 is not important.

Now apply exponent rule, distributing the power to both the 2 and the x where applicable.

Remember, multiplication is COMMUTATIVE, meaning we can multiply things in any order. So let's bring the numerical portion to the front of each term and multiply it all out.
