How do you factor #2x^2-x-15 #?

1 Answer
Aug 8, 2015

Answer:

#(2x + 5)(x - 3)#

Lol I hope you aren't trying to find your homework answers online to copy!

Explanation:

Firstly, consider the first term. Since it's #2x^2#, there are only two linear factors that it can be 'broken down' into: #x# and #2x#. Thus we can re-write the expression #2x^2 - x - 15# as #(2x + a)(x + b)# where #a# and #b# are real numbers to be determined

Thus
#2x^2 - x - 15 = (2x + a)(x + b)#
#2x^2 - x - 15 = 2x^2 + (a + 2b)x + ab#

Consider the last term. In order to get a negative number, we need one positive and one negative number. Thus, if #a# is positive, #b# is negative and vice versa.

Now lets think about the factors that make up #15#.
#15 = 1 * 15 = 3 * 5#

Now guess-and-check (not much to do) which combination (#1# and #15# or #3# and #5#) will give #-1# (coefficient of center term), that is, which #a# and #b# will fit the expression #a + 2b = -1#. Remember that if #a# is positive, #b# is negative and so on. With some luck you should get:

#5 + 2(-3) = -1#, #a = 5#, #b = -3#

Thus,
#2x^2 - x - 15 = (2x + a)(x + b) = (2x + 5)(x - 3)#