1.12 Completing the proof

Now $g_{1}(x)$ has $(s-1)$ distinct linear factors and $t$ distinct quadratic factors, so by induction we can express ${\frac{f(x)k(x)}{g_{1}(x)}}$ via partial fractions as in the statement of the Theorem. On the other hand, we have

for some $d$ and some real numbers $A_{0},\ldots,A_{d}$, and so

for some polynomial $p(x)$. Combining this with the expression for ${\frac{f(x)k(x)}{g_{1}(x)}}$ gives us the required expression for ${\frac{f(x)}{g(x)}}$. (If $g(x)$ has no linear factors then we perform the same trick with $g_{2}(x)$ and $Q_{t}(x)^{n_{t}}$, expressing $f(x)h(x)$ in the form $(B_{0}+C_{0}x)+\ldots+(B_{d}+C_{d}x)Q_{t}(x)^{d}$.) ∎