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Answer 1 · 2018-01-20T11:12:33

#2x^2-18=2(x+3)(x-3)#

First, we can bring out a #2# since both terms are a multiple of #2#:
#2x^2-18=2(x^2-9)#

Next we realize that #x^2-9# is a difference of squares. We know that the formula for difference of squares is:
#(a+b)(a-b)=a^2-b^2#

This means we can factor our expression like so:
#2(x+3)(x-3)#

Answer 2 · 2018-01-20T11:14:02

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

First, we know that #18=9*2#:
#2x^2-9*2#

Therefore, factor out the common term #2#:
#color(red)2x^2-9*color(red)2# to be equals to:

#2(x^2-9)#

Then, we rewrite #9# as #3^2#:

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

We will remove #2# for now.
We will then apply the DOS (Difference of squares) rule:
#x^2-y^2=(x+y)(x-y)#

Therefore, #(x^2-3^2)=(x+3)(x-3)#

Put back the #2#:
#2(x+3)(x-3)#

Answer 3 · 2018-01-20T11:15:14

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

It seems like there's nothing to be done here about the #x^2#. Let's start by factoring out the #2# to see if anything is "hidden".

#2x^2-18=2(x^2-9)#

#x^2-9=x^2-3^2=(x-3)(x+3)#

There's an identity for a difference of squares:

#x^2-y^2=(x-y)(x+y)#

On such problems, if "factoring out numbers" doesn't seem to work, maybe you'll have to use one of those.