Answer
The division gives a remainder of 10, meaning that $g(x)$ isn't a factor of $f(x)$.
Work Step by Step
$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space\space\space$ ${8x^2+5x+6}$
$x-1$$\space)\overline{8x^{3}-3x^{2}+x+4}$
$\space\space\space\space\space\space\space\space\space\space\space\space\underline{8x^{3}-8x^{2}}$
$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space5x^{2}+x$
$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\underline{5x^{2}-5x}$
$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space6x+4$
$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\underline{6x-6}$
$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space10$
Check answer:
$(8x^{2}+5x+6)\cdot(x-1)+10$
$8x^3-8x^2+5x^2-5x+6x-6+10$
$8x^{3}-3x^{2}+x+4\checkmark$
