Answer:
The potential roots are [tex]\dfrac{\pm 2}{3}, \dfrac{\pm 7}{2}\ and\ \pm 14 [/tex]
Step-by-step explanation:
We are given,
The function [tex]q(x)=6x^3+19x^2-15x-28[/tex].
Now, the factors of the constant 28 are 1,2,7,14 and 28.
And, the factors of co-efficient of highest degree i.e. 6 are 1,2,3 and 6.
So, dividing the factors of 28 by the factors of 6 gives us the potential roots,
[tex]\pm1, \pm 2,\pm 4,\pm 7,\pm 14,\pm 28,\pm \dfrac{1}{2}, \pm \dfrac{7}{2},\pm \dfrac{1}{3}, \pm \dfrac{2}{3},\pm \dfrac{4}{3}, \pm \dfrac{7}{3},\pm \dfrac{14}{3}, \pm \dfrac{28}{3},\pm \dfrac{1}{6},\pm \dfrac{7}{6}[/tex]
So, according to the options, we have,
