0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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? $\endgroup$ – lulu Oct 20 '17 at 12:38
  • 1
    $\begingroup$ 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 $0$. $\endgroup$ – lulu Oct 20 '17 at 12:41
  • $\begingroup$ 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? $\endgroup$ – Joey Oct 20 '17 at 12:47
  • $\begingroup$ 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 $r(x)=0$. $\endgroup$ – lulu Oct 20 '17 at 13:00
  • $\begingroup$ Think about the ordinary rational numbers. What's the remainder if you divide $\frac 12$ by $3$? $\endgroup$ – lulu Oct 20 '17 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.