jmc

We have $$f(x)=q(x)(x^2-1)+r(x),$$where $q(x)$ is the quotient and $r(x)$ is the remainder. Since $x^2-1$ is quadratic, the remainder is at most linear; let us write $r(x)=ax+b$.

Observe that $x=-1$ and $x=1$ are both zeroes of $x^2-1$. Thus $f(1)=r(1)$ and $f(-1)=r(-1)$.

We can use the given formula for $f(x)$ to compute $f(1)=-10$ and $f(-1)=16$. Thus we have the system of equations $$\{\begin{array}{l}-10=a\cdot (1)+b,\\ 16=a\cdot (-1)+b.\end{array}$$Adding these equations yields $6=2b$ and hence $b=3$. Substituting into either equation then yields $a=-13$.

Therefore, $r(x)=ax+b=\boxed{-13x+3}$.