Derivatives??? how to do a step

I havent taken a math class in a while and I was wondering how the got from paticular step in the book to another.


distance traveled/elapsed time =[ s(2+h)-s(2)]/[(2+h)-2]
= [16(2+h)^2-16(2)^2]/ h
=[16(4+4h+h^2)-16(4)]/h HOW DID THEY GET FROM THE LAST STEP TO THIS STEP?? ANY HELP APPRECIATED!!

[ 16 ( 2 + H ) ^2 - 16( 2 ) ^2] / h

( 2 + H ) ^2 = ( 2 + H ) * ( 2 + H ) = H^2 + 4h + 4 and the 2 ^ 2 = 4
so [ 16 ( H^2 + 4h + 4 ) - 16 (4 ) ] / h

first line: "s" must be equal to 16 . in the denominator if you take away the brackets it gives you just h.

second line: States new equation after simplifing once possible

third line:
(2+h)^2 = (2+h)(2+h) = 4 + 4h + h^2

2^2= 4

(2+h)^2 = (2+h)*(2+h) = 4 + 4h + h^2

(2)^2 = 2*2 = 4

substituing these two equations into the original equation gives

=[16(4+4h+h^2)-16(4)]/ h

They just substituted in (2+h)^2 = 4+4h+h^2 and (2)^2 = (4):

(2+h)^2 = (2+h)*(2+h)
= (2+h)*2 + (2+h)*h
= 2*2 + 2*h + 2*h + h*h
= 4 + 4h + h^2