Using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division

Here, $f(x)=3 x^{4}+2 x^{3}-x^{3} / 3-x / 9+2 / 27$

g(x) = x + 2/3

from remainder theorem when f(x) is divided by g(x) = x - (-2/3), the remainder is equal to f(- 2/3)

substitute the value of x in f(x)

$f\left(-\frac{2}{3}\right)=3\left(-\frac{2}{3}\right)^{4}+2\left(-\frac{2}{3}\right)^{3}-\frac{\left(-\frac{2}{3}\right)^{3}}{3}-\left[\frac{\left(-\frac{2}{3}\right)}{9}+\frac{22}{7}\left(\frac{2}{27}\right)\right]$

$=3\left(\frac{16}{81}\right)+2\left(\frac{-8}{27}\right)-\frac{4}{(9 * 3)}-\left(\frac{-2}{(9 * 3)}\right)+\frac{2}{27}$

$=\left(\frac{16}{27}\right)-\left(\frac{16}{27}\right)-\frac{4}{27}+\left(\frac{2}{27}\right)+\frac{2}{27}$

$=\left(\frac{4}{27}\right)-\left(\frac{4}{27}\right)$

= 0

Therefore, the remainder is 0