# How do you list all possible roots and find all factors of 6x^3+7x^2-3x-1?

Jan 17, 2017

The "possible" rational zeros are:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm 1$

The factorisation is:

$6 {x}^{3} + 7 {x}^{2} - 3 x - 1$

$= 3 \left(2 x - 1\right) \left(x + \frac{5}{6} - \frac{\sqrt{13}}{6}\right) \left(x + \frac{5}{6} + \frac{\sqrt{13}}{6}\right)$

#### Explanation:

Given:

$f \left(x\right) = 6 {x}^{3} + 7 {x}^{2} - 3 x - 1$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 1$ and $q$ a divisor of the coefficient $6$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{6} , \pm \frac{1}{3} , \pm \frac{1}{2} , \pm 1$

We find:

$f \left(\frac{1}{2}\right) = 6 \left(\frac{1}{8}\right) + 7 \left(\frac{1}{4}\right) - 3 \left(\frac{1}{2}\right) - 1 = \frac{3 + 7 - 6 - 4}{4} = 0$

So $x = \frac{1}{2}$ is a zero and $\left(2 x - 1\right)$ a factor:

$6 {x}^{3} + 7 {x}^{2} - 3 x - 1 = \left(2 x - 1\right) \left(3 {x}^{2} + 5 x + 1\right)$

We can factor the remaining quadratic by completing the square.

To reduce the need to do arithmetic with fractions, I will premultiply by $3 \cdot {2}^{2} = 12$ and divide by it at the end:

$12 \left(3 {x}^{2} + 5 x + 1\right) = 36 {x}^{2} + 60 x + 12$

$\textcolor{w h i t e}{12 \left(3 {x}^{2} + 5 x + 1\right)} = {\left(6 x\right)}^{2} + 2 \left(6 x\right) \left(5\right) + 25 - 13$

$\textcolor{w h i t e}{12 \left(3 {x}^{2} + 5 x + 1\right)} = {\left(6 x + 5\right)}^{2} - {\left(\sqrt{13}\right)}^{2}$

$\textcolor{w h i t e}{12 \left(3 {x}^{2} + 5 x + 1\right)} = \left(\left(6 x + 5\right) - \sqrt{13}\right) \left(\left(6 x + 5\right) + \sqrt{13}\right)$

$\textcolor{w h i t e}{12 \left(3 {x}^{2} + 5 x + 1\right)} = \left(6 x + 5 - \sqrt{13}\right) \left(6 x + 5 + \sqrt{13}\right)$

$\textcolor{w h i t e}{12 \left(3 {x}^{2} + 5 x + 1\right)} = 12 \cdot 3 \left(x + \frac{5}{6} - \frac{\sqrt{13}}{6}\right) \left(x + \frac{5}{6} + \frac{\sqrt{13}}{6}\right)$

So:

$3 {x}^{2} + 60 x + 12 = 3 \left(x + \frac{5}{6} - \frac{\sqrt{13}}{6}\right) \left(x + \frac{5}{6} + \frac{\sqrt{13}}{6}\right)$

Putting it all together:

$6 {x}^{3} + 7 {x}^{2} - 3 x - 1$

$= 3 \left(2 x - 1\right) \left(x + \frac{5}{6} - \frac{\sqrt{13}}{6}\right) \left(x + \frac{5}{6} + \frac{\sqrt{13}}{6}\right)$