How do you factor #n^2-8n+15#?

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Answer 1 · 2017-05-19T20:14:45

#(n - 3)(n - 5)#

First of all:

#(n +- x)(n +- y)#

We already have the first term, #n^2#.

Now, #x# times #y# needs to equal #15#. The factors or #15# are:

#1, 3, 5, 15#

Since the sum of #3# and #5# equal #8#, which is the coefficient of the middle term, #x = 3# and #y = 5# or vice versa.

Since the middle term is negative, both #x# and #y# are negative.

The final result is:

#(n - 3)(n - 5)#