Part $4$ of $6$

Test $x=\frac{1}{3}$

The remainder is

Therefore, $\frac{1}{3}$ a zero of $h(x).$

Part: $4/6$

Part $5$ of $6$

By the factor theorem, since $h(\frac{1}{3})=0,$ then $(x-\frac{1}{3})$ is a factor of $h(x).$

The quotient is also a factor of $h(x).$