Answer:
[tex]\frac{f(x)}{g(x)}=3x^2+6x-2-\frac{12}{3x+1}[/tex]
Step-by-step explanation:
The first function is [tex]f(x)=9x^3+21x^2-14[/tex]
The second function is [tex]g(x)=3x+1[/tex].
[tex]\frac{f(x)}{g(x)}=\frac{9x^3+21x^2-14}{3x+1}[/tex]
We perform the long division as shown in the attachment to obtain the quotient as: [tex]Q(x)=3x^2+6x-2[/tex] and remainder [tex]R=-12[/tex].
Therefore:
[tex]\frac{f(x)}{g(x)}=3x^2+6x-2-\frac{12}{3x+1}[/tex]
where [tex]x\ne -\frac{1}{3}[/tex]
