How do you factor #2x^2-x-3#?

1 Answer
EZ as pi
Aug 18, 2016

#(2x-3)(x+1)#

Explanation:

Combine the factors of 2 and 3 so by subtracting you get 1.
The middle term is #color(red)(1)x#

Write the factors and find the cross-products:
#color(white)(x.xx)(2)" "(3)#
#color(white)(xxxx)2" "3 rArr 1xx3 = 3#
#color(white)(xxxx)1" "1 rArr 2xx1 = 2" "3-2=1#

We have the correct factors, now sort out the signs:
We need to end up with -1. (more negatives)

#color(white)(xxxx)2" "-3 rArr 1xx-3 = -3#
#color(white)(xxxx)1" "+1 rArr 2xx+1 = +2" "-3+2=-1#

The top row is the first bracket and the bottom row is the second bracket.

#(2x-3)(x+1)#

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