How do you factor by grouping #m^2-n^2+5m-5n#?

1 Answer
Alan P.
Jun 18, 2018

#(m+n+5)(m-n)#

Explanation:

Group as #(color(red)(m^2-n^2))+(color(blue)(5m-5n))#

Recognize #(color(red)(m^2-n^2))# as the difference of squares which can be factored as #color(red)((m+n)(m-n))#

Extract the common factor of #color(blue)5# from #(color(blue)(5m-5n))# to get #color(blue)(5(m-n))#

Now we have
#(color(red)(m^2-n^2))+(color(blue)(5m-5n))=color(red)((m+n)(m-n))+color(blue)(5(m-n))#

Extracting the common factor of #(m-n)# from these two terms gives
#color(white)("XXX")(color(red)(m+n)+color(blue)5)(m-n)#

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