How do you find the product #(2q+5r)(2q-5r)#?

1 Answer
Mar 9, 2018

Answer:

#(2q+5r)(2q-5r)=color(red)(4q^2-25r^2)#

Explanation:

We know (hopefully you remember ...I know I do) that
#color(white)("XXX")(a+b)(a-b)=a^2-b^2# ...this is called "the difference of squares"

In this case using
#color(white)("XXX")2q# in place of #a#, and
#color(white)("XXX")5r# in place of #b#
we have
#(2q+5r)(2q-5r)=(2q)^2-(5r)^2#

#color(white)("XXXXXXXXXXX")=4q^2-25r^2#

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Of course, if you don't remember "the difference of squares", you could apply the distributive property:
#(2q+5r)(2q-5r)#
#color(white)("XXX")=2q(2q-5r)+5r(2q-5r)#
#color(white)("XXX")=(2q)^2-(2q)(5r)+(5r)(2q) - (5r)^2#
#color(white)("XXX")=(2q)^2-(5r)^2#
#color(white)("XXX")=4q^2-25r^2#