How do you find the product of #(7u+4v)(7u-4v)#?

1 Answer
Jan 21, 2017

Answer:

#(7u+4v)(7u-4v) = 49u^2-16v^2#

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 #a=7u# and #b=4v#

Hence:

#(7u+4v)(7u-4v) = (7u)^2-(4v)^2 = 49u^2-16v^2#