Answer
$\text{improper}$;
$3+\dfrac{1}{x^{2}-1}$
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 $A(x)= 3x^2-2$ is $2$ and the degree of the denominator ; $B(x)=x^2-1$ is $2$.
We see that the given rational expression is improper.
To make the rational expression proper, we will solve as:
$ \dfrac{3x^{2}-2}{x^{2}-1}=\dfrac{3x^{2}-3+1}{x^{2}-1} \\
\\=\dfrac{3x^{2}-3}{x^{2}-1}+\dfrac{1}{x^{2}-1}
\\=\dfrac{3(x^{2}-1)}{x^{2}-1}+\dfrac{1}{x^{2}-1}
\\=3+\dfrac{1}{x^{2}-1}$
