How do you multiply #(2y + 11) ( 2y - 11)#?

1 Answer
May 28, 2017

#4y^2-121#

Explanation:

If you didn't already know, we can use the #"F.O.I.L"# method.
#"F.O.I.L"# stands for:

#color(red)"F"#: First, as in you multiply the first two terms in each #()#
#color(blue)"O"#: Outer (multiply the outer terms)
#color(green)"I"#: Inner (multiply the inner terms)
#color(orange)"L"#: Last (multiply the last terms

#(color(red)(2y)+11)(color(red)(2y)-11) : color(red)(4y^2)#

#(color(blue)(2y)+11)(2ycolor(blue)(-11)) : color(blue)(-22y)#

#(2y+color(green)(11))(color(green)(2y)-11) : color(green)(22y)#

#(2y+color(orange)(11))(2ycolor(orange)(-11)) : color(orange)(-121)#

Gathering this information, we can rewrite the expression and combine like terms:

#4y^2color(red)(-22y+22y)-121#

#4y^2cancel(color(red)(-22y+22y))-121#

The middle terms cancel and so we are left with this as our final answer:

#4y^2-121#