How do you factor the trinomial #x^2-4x-12#?

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Answer 1 · 2015-11-19T16:28:27

#f(x) = (x-6)(x+2)#

#f(x)=ax^2+bx+c#

All you have to do is find two numbers which added give you #b# and multiplied give you #c#.

Those numbers are #-6# and #2#.

#f(x) = (x-6)(x+2)#

#f(x) = x^2+2x-6x-12#

#f(x) = x^2-4x-12#

Answer 2 · 2015-11-19T16:40:31

#(x-6)(x+2) #

Have a look at the explanation

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The trick is to spot what value for a and b give you

#ab=12# We need -12 so one of a and b is negative and the other positive

#a-b =4# We need -4 so the larger of a and b is negative

Consider the factors of 12

#color(red)(1 times 12" and the difference is 11 so it fails")#

#color(green)(2 times 6 " and the difference is 4 so it works")#

#color(red)(3 times 4" and the difference is 1 so it fails")#
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Try 1
#(x + 6)(x - 2) # The larger being positive which is the wrong way
#color(white)(xxxxxxxxxx)#round to what we stated

#= x^2 -2x +6x -12# This fails as it gives #+4x#

Try2
#(x-6)(x+2) #

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

# = x^2-4x-12# This works so it is the answer!