Home > Algebra > Factoring by Grouping

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

Answer 1

Ethan Adkins

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

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)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Nathan Axel

To factor by grouping the expression ( m^2 - n^2 + 5m - 5n ), first, group the terms with common factors together. Then, factor out the greatest common factor from each group.

This yields:

( (m^2 + 5m) - (n^2 + 5n) )

Now, factor out the common factors from each group:

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

Now, you can see that both terms have a common factor of ( (m + 5) ), so factor that out:

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

Therefore, the factored form of the expression is ( (m + 5)(m - n) ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.