Answer:
Step-by-step explanation:
3[tex]x^{3}[/tex][tex](2y)^{2}[/tex]4[tex]x^{4}[/tex]
So I wrote down exactly above how the eq. is put in the question, note that the 2y is the only coefficient that is inside of the parenthesis, I hope that is how the question is asked. The 3 and 4 coefficient are NOT part of the exponential part.. is my point.. which will make a big difference if they were.
anyway
3[tex]x^{3}[/tex]4[tex]y^{2}[/tex]4[tex]x^{4}[/tex]
48[tex]y^{2}[/tex][tex]x^{3}[/tex][tex]x^{4}[/tex] ( b/c the x bases are the same we can just add the exponents)
48[tex]y^{2}[/tex][tex]x^{3+4}[/tex]
48[tex]y^{2}[/tex][tex]x^{7}[/tex] ( this is your answer )
side note: an important learning point that was implied above is that you could rewrite the expression as 48[tex]y^{2}[/tex][tex]x^{2}[/tex][tex]x^{2}[/tex][tex]x^{2}[/tex][tex]x^{1}[/tex]... why would you want to write it that way? b/c often a square is needed to make a problem work out easier.
as in Cos^2(Ф) + Sin^2(Ф) = 1 but that's another question L :P
(if you're read this far.. good for you.. if it's the 1st time you've heard of that rule for exponents then ask your teacher / professor if I can have 1/2 of their pay since I am doing their job :D )
