#### Answer

$\text{improper}$;
$3x+\dfrac{x^2-24x-2}{x^{3}+8}$

#### Work Step by Step

A rational expression $\dfrac{A(x)}{B(x)}$ is said to be proper if the degree of polynomial in numerator is less than the degree of its denominator.If it does not happen , then it is said to be improper rational polynomial.
Here, the degree of the numerator $A(x)= 3x^4+x^2-21$ is $4$ and the degree of the denominator ; $B(x)=x^3+8$ is $3$.
We see that the given rational expression is improper.
To make the rational expression proper, we will solve as:
$ \dfrac{3x^4+x^2-2}{x^{3}+8}=\dfrac{3x^{4}+24x+x^2-24x-2}{x^{3}+8}
\\=\dfrac{(3x^{4}+24x)+(x^2-24x-2)}{x^{3}+8}
\\=\dfrac{3x(x^{3}+8)+(x^2-24x-2)}{x^{3}+8}
\\=\dfrac{3x(x^{3}+8)}{x^{3}+8}+\dfrac{x^2-24x-2}{x^{3}+8}
\\=3x+\dfrac{x^2-24x-2}{x^{3}+8}$