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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 =
\(\rm \_\_(2x)^34^0+\_\_(2x)^24^1+\_\_(2x)^14^2+\_\_(2x)^04^3\)
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:
\(\rm 1(2x)^34^0+3(2x)^24^1+3(2x)^14^2+1(2x)^04^3\)
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.
\(\rm (2x)^3+3(2x)^24+3(2x)4^2+4^3\)
Now apply exponent rule, distributing the power to both the 2 and the x where applicable.
\(\rm 2^3x^3+3\cdot2^2x^24+3\cdot2x4^2+4^3\)
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.
\(\rm 2^3x^3+3\cdot2^2\cdot4x^2+3\cdot2\cdot4^2x+4^3\)
\(\rm 8x^3+48x^2+96x+64\)
