How do you factor the trinomial #n^2+4n-21#?

2 Answers
Sep 6, 2017

We need to play with factors of #21# (#+-1, +-3, +-7, +-21#) which add to #+4#:

#n^2 + 4n - 21 => (n + 7)(n - 3)#

Sep 6, 2017

#n^2+4n-21 = (n-3)(n+7)#

Explanation:

Note that #(n+2)^2 = n^2+4n+4#, which matches the given expression in the first two terms.

So we find:

#n^2+4n-21 = n^2+4n+4-25#

#color(white)(n^2+4n-21) = (n+2)^2-5^2#

#color(white)(n^2+4n-21) = ((n+2)-5)((n+2)+5)#

#color(white)(n^2+4n-21) = (n-3)(n+7)#