Question #eca0b

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Question #eca0b

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Answer 1 · 2016-08-16T21:51:37

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 $3x-1$ :
$(3x-1)(x^2+6)$

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Question #eca0b

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