# Why does the reciprocal of a polynomial divided by another polynomial not have a remainder?

From a question in my textbook:
Given that $\,f(x) = 8x^3 + 4x -3$, find the remainder, if it exists, when $\frac1{\,f(x)}$ is divided by $x + 1$.

The answer states that the remainder does not exist. Does this mean that the reciprocal of any polynomial does not have a remainder when divided with another polynomial? If that's the case, why?

• Well...what would it mean? when you divide one polynomial, $g(x)$ by another, $h(x)$ you write $g(x)=h(x)\times q(x)+r(x)$ where $q(x)$, the quotient, and $r(x)$, the remainder are both polynomials and the degree of $r(x)$ is less than the degree of $h(x)$. How would you define it for rational functions? – lulu Oct 20 '17 at 12:38
• You could make a case that the remainder is $0$, as we can obviously write $\frac 1{f(x)}=(x+1)\times q(x)+0$ where $q(x)=\frac 1{(x+1)f(x)}$ is rational. That's not a very interesting construction, though. The remainder would always be $0$ as the reciprocal of a rational function is again rational. Just as the remainder on dividing one rational number by another is always $0$. – lulu Oct 20 '17 at 12:41
• If the remainder were to be written as $r(x)$, then wouldn't $\frac1{(x+1)\,f(x)} = q(x) + \frac{r(x)}{x+1}$, and wouldn't there still be a remainder? – Joey Oct 20 '17 at 12:47
• What's $q(x)$ meant to be? If we allow $q(x)$ to be a rational function then just take $q(x)=\frac 1{(x+1)f(x)}$ so $r(x)=0$. – lulu Oct 20 '17 at 13:00
• Think about the ordinary rational numbers. What's the remainder if you divide $\frac 12$ by $3$? – lulu Oct 20 '17 at 13:01