Answer:
[tex]\frac{x+4}{x} \hspace{8}where\hspace{8}x\neq0,\frac{2}{3}[/tex]
Step-by-step explanation:
The division between functions is defined as:
[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)} ,\hspace{10}g(x)\neq0[/tex]
So:
[tex](\frac{f}{g} )(x)=\frac{3x^2+10x-8}{3x^2-2x}\\ \\Factor\hspace{3}x\hspace{3}out\hspace{3}the\hspace{3}denominator\\\\(\frac{f}{g} )(x)=\frac{3x^2+10x-8}{x(3x-2)}\\\\Factor\hspace{3}the\hspace{3}numerator\\\\(\frac{f}{g} )(x)=\frac{4(3x-2)+x(3x-2)}{x(3x-2)}\\\\Factor\hspace{3}3x-2\hspace{3}from\hspace{3}the\hspace{3}numerator\\\\(\frac{f}{g} )(x)=\frac{(3x-2)(x+4)}{x(3x-2)}=\frac{x+4}{x}[/tex]
Since [tex]g(x) \neq0[/tex] , let's find its roots:
[tex]3x^2-2x=0\\\\Factor\\\\x(3x-2)=0\\\\Split\hspace{3}into\hspace{3}two\hspace{3}equations:\\\\(1):x=0\\(2):3x-2=0[/tex]
For (2)
[tex]3x=2\rightarrow x=\frac{2}{3}[/tex]
Therefore the roots are:
[tex]x=0,\hspace{3}x=\frac{2}{3}[/tex]
Finally the complete answer is:
[tex](\frac{f}{g})(x)= \frac{x+4}{x} \hspace{8}where\hspace{8}x\neq0,\frac{2}{3}[/tex]
