Question #eca0b

1 Answer
Aug 16, 2016

Answer:

Do some factoring to get #3x^3-x^2+18x-6=(3x-1)(x^2+6)#

Explanation:

First we check to make sure the ratios are the same for consecutive terms. In plain English, this means we do this:
#color(blue)(3x^3)-color(blue)(x^2)+color(red)(18x)-color(red)(6)#
#->color(red)(6/(18x))=1/(3x)#
#->color(blue)(x^2/(3x^2))=1/(3x)#

Because the ratios are the same, we can factor by grouping.

Now, let's pull an #x^2# out of #3x^3-x^2#:
#3x^3-x^2+18x-6#
#->x^2(3x-1)+18x-6#

And a #6# out of #18x-6#:
#x^2(3x-1)+18x-6#
#->x^2(3x-1)+6(3x-1)#

Note that these have a common term of #(3x-1)#:
#x^2color(red)((3x-1))+6color(red)((3x-1))#

That means we can pull out a #3x-1# also:
#x^2color(red)((3x-1))+6color(red)((3x-1))#
#->color(red)((3x-1))(x^2+6)#

This last part may seem confusing. If it helps, replace #3x-1# with something less intimidating, like #a#:
#x^2a+6a#

For me, it's easier to see that we can pull out an #a# as a common factor:
#x^2a+6a#
#->a(x^2+6)#

Now just replace #a# with #3x-1#:
#(3x-1)(x^2+6)#