Answer:
[tex]\pm 1,\pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \dfrac{1}{2},\pm \dfrac{3}{2},\pm \dfrac{1}{3},\pm \dfrac{2}{3},\pm \dfrac{4}{3},\pm \dfrac{1}{6}.[/tex]
Step-by-step explanation:
According to the rational root theorem, the all possible rational roots of a polynomial are in the form of [tex]\pm \dfrac{p}{q}[/tex], where p is a factor of constant and q is a factor of leading coefficient.
The given polynomial is
[tex]S(x)=6x^4-x^2+3x+12[/tex]
Here, leading term is 6 and constant is 12.
Factors of 6 are ±1, ±2, ±3 and ±6.
Factors of 12 are ±1, ±2, ±3, ±4, ±6 and ±12.
Using rational root theorem, the all possible rational roots of given polynomial is
[tex]\pm 1,\pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \dfrac{1}{2},\pm \dfrac{3}{2},\pm \dfrac{1}{3},\pm \dfrac{2}{3},\pm \dfrac{4}{3},\pm \dfrac{1}{6}.[/tex]
