# Question #eca0b

Aug 16, 2016

Do some factoring to get $3 {x}^{3} - {x}^{2} + 18 x - 6 = \left(3 x - 1\right) \left({x}^{2} + 6\right)$

#### Explanation:

First we check to make sure the ratios are the same for consecutive terms. In plain English, this means we do this:
$\textcolor{b l u e}{3 {x}^{3}} - \textcolor{b l u e}{{x}^{2}} + \textcolor{red}{18 x} - \textcolor{red}{6}$
$\to \textcolor{red}{\frac{6}{18 x}} = \frac{1}{3 x}$
$\to \textcolor{b l u e}{{x}^{2} / \left(3 {x}^{2}\right)} = \frac{1}{3 x}$

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

Now, let's pull an ${x}^{2}$ out of $3 {x}^{3} - {x}^{2}$:
$3 {x}^{3} - {x}^{2} + 18 x - 6$
$\to {x}^{2} \left(3 x - 1\right) + 18 x - 6$

And a $6$ out of $18 x - 6$:
${x}^{2} \left(3 x - 1\right) + 18 x - 6$
$\to {x}^{2} \left(3 x - 1\right) + 6 \left(3 x - 1\right)$

Note that these have a common term of $\left(3 x - 1\right)$:
${x}^{2} \textcolor{red}{\left(3 x - 1\right)} + 6 \textcolor{red}{\left(3 x - 1\right)}$

That means we can pull out a $3 x - 1$ also:
${x}^{2} \textcolor{red}{\left(3 x - 1\right)} + 6 \textcolor{red}{\left(3 x - 1\right)}$
$\to \textcolor{red}{\left(3 x - 1\right)} \left({x}^{2} + 6\right)$

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

For me, it's easier to see that we can pull out an $a$ as a common factor:
${x}^{2} a + 6 a$
$\to a \left({x}^{2} + 6\right)$

Now just replace $a$ with $3 x - 1$:
$\left(3 x - 1\right) \left({x}^{2} + 6\right)$