# What are the possible rational roots of 6x^4+35x^3-x^2-7x-1=0 and then determine the rational roots?

Jan 9, 2017

The possible rational roots are +-1;+-1/2;+-1/3;+-1/6;
The rational roots are 1/2;-1/3

#### Explanation:

The possible rational roots are obtained by dividing each of the factors of the known term (-1) by each of the factors of the maximum degree term's coefficient (6).

Then they are:

+-1;+-1/2;+-1/3;+-1/6

To determine the rational roots, let's use the remainder rule:

$\textcolor{red}{x = 1} \to 6 + 35 - 1 - 7 - 1 = 32 \ne 0$
$\textcolor{red}{x = - 1} \to 6 - 35 - 1 + 7 - 1 = - 24 \ne 0$
$\textcolor{red}{x = \frac{1}{2}} : 6 {\left(\frac{1}{2}\right)}^{4} + 35 {\left(\frac{1}{2}\right)}^{3} - {\left(\frac{1}{2}\right)}^{2} - 7 \left(\frac{1}{2}\right) - 1 = 0$

Then $x = \frac{1}{2}$ is a rational root.

You also can use a spreadsheet to find zeros faster.

Then you would find only another rational root $x = - \frac{1}{3}$