How do you factor #n^2+4n-12#?

2 Answers
Mar 28, 2018

#(n-2)(n+6)#

Explanation:

By using SUM PRODUCT

= #n^2+6n-2n-12#

= #n(n+6)-2(n+6)#

= #(n-2)(n+6)#

Hope this helps!

Mar 28, 2018

#(n+6)(n-2)#

Explanation:

To factor this we, have to split the middle term.
If the quadratic equation is #ax^2+bx+c#, then we need to split the #bx# into two terms such that the ratio of #a# to first half = second half to #c#

So, we split #n^2+4n-12# into #n^2+6n-2n-12#
As we can see, #1:6#=#-2:-12#

Now in the first and second half take common the biggest term possible common

=#(n+6)n-(n+6)2#

Here take a look, if the terms inside brackets are same, then you're on the right track

Now take the remaining outside the bracket common and you get

#(n+6)(n-2)#