How do you write #4x^2+2x-12# in factored form?

1 Answer
Sep 16, 2015

Answer:

#4x^2+2x-12#
#color(white)("XXXXXX")=color(green)(2)color(red)((2x+3))color(blue)((x-2))#

Explanation:

Extract the obvious constant factor of #2# to simplify

#color(green)(2)color(orange)((2x^2+x-6))#

Factoring
by looking for integer factors #a# and #b# of #2#
and integer factors #c# and #d# of #(-6)#
such that
#(ad+bc)x = 1x#

The only integer factors of #2# are #1xx2#

Integer factors of #(-6)# are #{(1xx-6),(2xx-3)(-1xx6),(-3xx2)}#

Checking the four possible combinations, we find:
#color(orange)((2x^2+x-6))=color(red)((2x+3))color(blue)((x-2))#

So #4x^2+2x-12#
#color(white)("XXX")=color(green)(2)color(orange)((2x^2+x-6))#
#color(white)("XXX")=color(green)(2)color(red)((2x+3))color(blue)((x-2))#