How do you factor the expression #2x^2 - 11x + 12#?

1 Answer
Jan 28, 2016

Answer:

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

Explanation:

Looking at the #x^2# coefficient, #2#, we know that the coefficients of #x# in each factor can only be #1# and #2# since no other integers multiply to 2.

Therefore we can set up the factors as:
#(2x - "__")*(x - "__")#

We know that both of the operations in the factors must be subtraction because #12# is positive and #-11# is negative.

To find out what goes in the blanks, we need to check the factors of 12:

#1# and #12#
#2# and #6#
#3# and #4#

#1# and #12# can't work, because no matter what side each is on, #12# times anything will be greater than #11#.

#2*2 + 1*6# is #10#, and #2*6 + 1*2# is #14#, neither of which are 11.

Finally, we can try pair #3# and #4#. Feel free to try both of them yourself to see why the first must be #3# and the second must be #4#, not the other way around!