How do you factor #8x^2 - 4x - 24#?

2 Answers
Aug 21, 2017

Well, you can factor out 4 immediately:

#4(2x^2 - x - 6)#

You can work out that the further factorization will look something like:

#(2x + a)(x + b)# where #a*b = -6#

and

#(2x * b) + (a * x) = -1x#

so you know that #2a + b= -1#
and

#a*b = -6#

You know that either a or b must be negative, and the other number must be positive.

At this point, I usually resort to trial and error guessing. There are usually a couple of choices, so try one, then the other.

so what if #a = 3# and #b = -2#, does that work?

#(2x + 3)(x - 2) = 2x^2 -4x + 3x -6 = 2x^2 - x -6#

...good guess. So your factorization will be:

#4(2x + 3)(x - 2)#

Aug 21, 2017

#=4(2x+3)(x-2)#

Explanation:

First take out the common factor of #4#

#8x^2 -4x -24#

#=4(2x^2 -x-6)#

Find factors of #2 and 6# whose products differ by #1#

#" "2 and 6#
#" "darr" "darr#

#" "2" "3" "rarr 1 xx 3 = 3#
#" "1" "2" "rarr 2xx2 = ul4#
#color(white)(xxxxxxxxx.x.xxxxxx)1#

#8x^2 -4x -24#

#=4(2x^2 -x-6)#

#=4(2x+3)(x-2)#