Is #a^2-b^2# equal to #(a-b)(a+b)#?

2 Answers
Mar 21, 2018

Yes!

This is one that is really worth committing to memory. It crops up quite often.

Explanation:

Given:

#color(blue)((a-b))color(green)( (a+b) )#

Multiply everything in the right bracket by everything in the left

#color(green)(color(blue)(a)(a+b) color(white)("dd")color(blue)(-b)(a+b))# Notice the way the minus follows the #b#

#a^2+abcolor(white)("dd")-ba+b^2#

But #-ba# has exactly the same value as #-ab# so we can write:

#a^2+cancel(ab) color(white)("dd")-cancel(ab)+b^2#

#a^2+b^2#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus #a^2+b^2=(a-b)(a+b)#

Mar 21, 2018

#"yes"#

Explanation:

#a^2-b^2" is called a "color(blue)"difference of squares"#

#"since "a^2" and "b^2" are perfect squares separated by a"#
#"difference (-)"#

#"to verify this result expand the factors"#

#rArr(color(red)(a-b))(a+b)#

#=color(red)(a)(a+b)color(red)(-b)(a+b)larrcolor(blue)"distribute"#

#=a^2cancel(+ab)cancel(-ab)-b^2#

#=a^2-b^2#

#"Useful factoring tool " color(red)(bar(ul(|color(white)(2/2)color(black)(a^2-b^2=(a-b)(a+b))color(white)(2/2)|)))#