Answer: [tex]\bold{D.\ \dfrac{x+4}{x}\ where\ x \neq 0, \dfrac{2}{3}}[/tex]
Step-by-step explanation:
f(x) = 3x² + 10x - 8
g(x) = 3x² - 2x
[tex]\dfrac{f(x)}{g(x)}=\dfrac{3x^2+10x-8}{3x^2-2x}=\dfrac{(3x-2)(x+4)}{x(3x-2)}=\dfrac{x+4}{x}[/tex]
Restriction on x: the denominator cannot be equal to zero. So,
x(3x - 2) ≠ 0
→ x ≠ 0 and 3x - 2 ≠ 0
3x ≠ 2
x [tex]\neq \dfrac{2}{3}[/tex]
