# How do you factor y= 6x^3+13x^2-14x+3 ?

Mar 21, 2018

$\textcolor{red}{\left(2 x - 1\right) \left(3 x - 1\right) \left(x + 3\right)}$

#### Explanation:

We can optimistically hope that $y = \textcolor{\lim e}{6} {x}^{3} + 13 {x}^{2} - 14 x + \textcolor{m a \ge n t a}{3}$
has at least one rational root.

By the Rational Root Theorem we know that any such root must be of the form $\textcolor{b l u e}{\frac{p}{q}}$
where $\textcolor{b l u e}{p}$ is an integer factor of $\textcolor{m a \ge n t a}{3}$ (for this expression)
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{p} \in \left\{\pm 1 , \pm 3\right\}$
and
$\textcolor{b l u e}{q}$ is an integer factor of $\textcolor{\lim e}{6}$ (again, for this expression)
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{q} \in \left\{\pm 1 , \pm 2 , \pm 3 , \pm 6\right\}$

There turn out to only be 10 possible rational roots using this information:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{\frac{p}{q}} \in \left\{\pm 3 , \pm 1 , \pm \frac{1}{2} , \pm \frac{1}{3} , \pm \frac{1}{6}\right\}$

You could evaluate these manually, but I used a spreadsheet: In this case, we have been extremely lucky, since we have all 3 possible roots identified $\left\{0.5 = \frac{1}{2} , 0.333 \overline{3} = \frac{1}{3} , - 3\right\}$

Which implies the factors:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{\lim e}{6} \cdot \left(x - \frac{1}{2}\right) \cdot \left(x - \frac{1}{3}\right) \cdot \left(x + 3\right)$
or (after using the $\textcolor{\lim e}{6}$ to clear the fractions)
$\textcolor{w h i t e}{\text{XXX}} \left(2 x - 1\right) \left(3 x - 1\right) \left(x + 3\right)$