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

1 Answer
May 28, 2015

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

Problem: Factor #2x^2-x-3#.

The generic form of this equation is #ax^2+bx+c#.

#a=2#
#b=-1#
#c=-3#

Multiply #a# and #c#.

#2*(-3)=-6#

Find two factors of #-6# that when added equal #-1#. The numbers #-3# and #2# fit this requirement.

Rewrite the equation so that #-3x# and #2x# replace #-1x#.

Group the first and second pairs of terms.

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

Factor #x# out of the first term.

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

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

Factor out the common term #2x-3#.

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

We can also rewrite the equation as #2x^2+2x-3x-3#.

Group the two sets of terms.

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

Factor #2x# from the first term, and #3# out of the second term.

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

Factor out the common term #x+1#.

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