How do you factor #n^2+8n=12#?

1 Answer
May 28, 2015

To be technically accurate (at least as I was taught) you factor expressions not equations.
I will assume that what you really want to do is find solutions for the given equation #n^2+8n=12#
(and then re-write the equation as a pair of factors equal to zero).

There are several ways to approach this:
trial and error;
quadratic formula for roots; or
completion of the square.

By completion of the square
#n^2+8n = 12#

#rarr n^2+8n+4^2 = 12+ 4^2#

#rarr (n+4)^2 = 28#

#rarr (n+4) = +-sqrt(28)#

#rarr n= -4 +-2sqrt(7)#

Re-expressing the original equation as a pair of factors equal to zero
#(n+4+2sqrt(7))(n+4-2sqrt(7)) = 0#