How do you find the product of #(7u+4v)(7u-4v)#?
1 Answer
Jan 21, 2017
Explanation:
You can use the FOIL mnemonic if it helps...
#(7u+4v)(7u-4v) = overbrace((7u)(7u))^"First" + overbrace(color(red)(cancel(color(black)((7u)(-4v)))))^"Outside" + overbrace(color(red)(cancel(color(black)((4v)(7u)))))^"Inside" + overbrace((4v)(-4v))^"Last"#
#color(white)((7u+4v)(7u-4v)) = 49u^2-16v^2#
Alternatively, spot that this is a difference of squares:
#(a+b)(a-b) = a^2-b^2#
with
Hence:
#(7u+4v)(7u-4v) = (7u)^2-(4v)^2 = 49u^2-16v^2#