How do you factor the expression #3x^2 - 6x - 9#?

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Answer 1 · 2018-03-10T18:46:46

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

#3x^2-6x-9#
#=3(x^2-2x-3)# (taking out the common factor of 3)
#=3(x+1)(x-3)# ("looking for factors of 3 to give -2")

Answer 2 · 2018-03-10T18:50:27

#x=3# and #x=-1#

All terms have a #3# in common, so we can factor this out. When we factor out a #3#, we are essentially dividing every term by #3#. We get:

#3(x^2-2x-3)=0#

Now, we can factor the inside, if we think of two numbers, when I add them, add up to #-2#, and those same two numbers have a product of #-3#.

#-3+1=2#, and #-3*1=-3#, therefore #-3# and #1# are our two factors. We get:

#3(x-3)(x+1)=0#

We can simplify this further by setting every term equal to zero and we get two solutions:

#x=3# and #x=-1#