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How do you factor by grouping $m^{2} - n^{2} + 5 m - 5 n$ $m^2-n^2+5m-5n$ ?

1 Answer

Jun 18, 2018

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

Explanation:

Group as $(m^{2} - n^{2}) + (5 m - 5 n)$ $(color(red)(m^2-n^2))+(color(blue)(5m-5n))$

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

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

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

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

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