Suppose that $f(x)=x^{4}-x^{3}-1$ and $g(x)=x^{8}-x^{6}-2 x^{4}+1$. If $g(x)=f(x) h(x)$, determine the polynomial function $h(x)$.
Since $f(x)$ has degree $4, g(x)$ has degree 8 , and $h(x)$ is a polynomial with $g(x)=f(x) h(x)$, then $h(x)$ has degree 4 .

We write $h(x)=a x^{4}+b x^{3}+c x^{2}+d x+e$ and so

$$
x^{8}-x^{6}-2 x^{4}+1=\left(x^{4}-x^{3}+0 x^{2}+0 x-1\right)\left(a x^{4}+b x^{3}+c x^{2}+d x+e\right)
$$

Comparing coefficients of $x^{8}$ on the left and right sides, we obtain $1=1 \cdot a$ and so $a=1$, giving

$$
x^{8}-x^{6}-2 x^{4}+1=\left(x^{4}-x^{3}+0 x^{2}+0 x-1\right)\left(x^{4}+b x^{3}+c x^{2}+d x+e\right)
$$

Comparing coefficients of $x^{7}$, we obtain $0=1 \cdot b+(-1) \cdot 1$ and so $b=1$, giving

$$
x^{8}-x^{6}-2 x^{4}+1=\left(x^{4}-x^{3}+0 x^{2}+0 x-1\right)\left(x^{4}+x^{3}+c x^{2}+d x+e\right)
$$

Comparing coefficients of $x^{6}$, we obtain $-1=1 \cdot c+(-1) \cdot 1+0 \cdot 1$ and so $c=0$, giving

$$
x^{8}-x^{6}-2 x^{4}+1=\left(x^{4}-x^{3}+0 x^{2}+0 x-1\right)\left(x^{4}+x^{3}+0 x^{2}+d x+e\right)
$$

Comparing coefficients of $x^{5}$, we obtain $0=1 \cdot d+(-1) \cdot 0+0 \cdot 1+0 \cdot 1$ and so $d=0$, giving

$$
x^{8}-x^{6}-2 x^{4}+1=\left(x^{4}-x^{3}+0 x^{2}+0 x-1\right)\left(x^{4}+x^{3}+0 x^{2}+0 x+e\right)
$$

Comparing constant terms, we obtain $1=(-1) \cdot e$ and so $e=-1$, giving

$$
x^{8}-x^{6}-2 x^{4}+1=\left(x^{4}-x^{3}+0 x^{2}+0 x-1\right)\left(x^{4}+x^{3}+0 x^{2}+0 x-1\right)
$$

We can verify by expansion that the remaining terms match.

Therefore, $h(x)=x^{4}+x^{3}-1$.

This result can also be obtained by polynomial long division.

ANSWER: $h(x)=x^{4}+x^{3}-1$