1) We can do this by listing all the factors of -1, and the leading coefficient 6. So, we can write them as a ratio this way:
[tex]\frac{p}{q}=\pm\frac{1}{1,\:2,\:3,\:6}[/tex]Note that p stands for the constant and q the factors of that leading coefficient
2) Now, let's test them by plugging them into the polynomial. If it is a rational root it must yield zero:
[tex]\begin{gathered} 6x^3+7x^2-3x+1=0 \\ 6(\pm1)^3+7(\pm1)^2-3(\pm1)+1=0 \\ 71\ne0,5\ne0 \\ \frac{1}{2},-\frac{1}{2} \\ 6(\pm\frac{1}{2})^3+7(\pm\frac{1}{2})^2-3(\pm\frac{1}{2})+1=0 \\ 2\ne0,\frac{7}{2}\ne0 \\ \\ 6(\pm\frac{1}{3})^3+7(\pm\frac{1}{3})^2-3(\pm\frac{1}{3})+1=0 \\ 1\ne0,\frac{23}{9}\ne0 \\ \frac{1}{6},-\frac{1}{6} \\ 6(\frac{1}{6})^3+7(\frac{1}{6})^2-3(\frac{1}{6})+1=0 \\ \frac{13}{18}\ne0,-\frac{5}{3}\ne0 \end{gathered}[/tex]3) So the possible roots are:
[tex]\pm1,\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{6}[/tex]But there are no actual rational roots.
