Answer
$f(-\frac{1}{3})=4$.
Work Step by Step
The given equation is $f(x)=3x^4+7x^3+8x^2+2x+4$.
If the given equation is divided by $x+\frac{1}{3}$ the remainder is $f(-\frac{1}{3})$.
Use synthetic division to divide.
Use synthetic division
$\begin{matrix}
-\frac{1}{3}) &3&7&8&2&4 \\
& &-1&-2&-2 &0\\
& --&--&--& --&--\\
& 3&6&6&0&4
\end{matrix}$
The remainder, $4$, is the value of $f(-\frac{1}{3})$.
Check:
$\Rightarrow f(-\frac{1}{3})=3(-\frac{1}{3})^4+7(-\frac{1}{3})^3+8(-\frac{1}{3})^2+2(-\frac{1}{3})+4$.
Simplify.
$\Rightarrow f(-\frac{1}{3})=\frac{1}{27}-\frac{7}{27}+\frac{8}{9}-\frac{2}{3}+4$.
The LCD is $27$.
Multiply the numerator and the denominator:
$\Rightarrow f(-\frac{1}{3})=\frac{1}{27}-\frac{7}{27}+\frac{24}{27}-\frac{18}{27}+\frac{108}{27}$.
Add numerators because denominators are the same.
$\Rightarrow f(-\frac{1}{3})=\frac{1-7+24-18+108}{27}$.
Simplify.
$\Rightarrow f(-\frac{1}{3})=\frac{108}{27}$.
$\Rightarrow f(-\frac{1}{3})=4$. True.
