Generally: Suppose we are given a polynomial of arbitrary degree $P(x)$ with arbitrary coefficients $a_1,a_2,\cdots,a_n$, and we divide $P(x)$ by $(x-r_1)$ where $r_1$ is one of its roots, should we expect a remainder when performing synthetic division?

If there is a remainder, say $3a_2+a_1+3$, should this be set to equal $0$?

  • $\begingroup$ No why would we? Since $x=r$ is a root, $(x-r)$ is a factor! $\endgroup$ – King Tut Apr 3 '18 at 17:43
  • 1
    $\begingroup$ Are you expecting a different result from synthetic division than from any other form of division? $\endgroup$ – robjohn Apr 3 '18 at 17:43
  • $\begingroup$ We have no idea what the coefficients are, we are just given a root. And there appears to be a remainder after synthetic division. Am I just making an arithmetic error in my division? @KingTut $\endgroup$ – John Glenn Apr 3 '18 at 17:44
  • $\begingroup$ You might consider showing the actual problem and the work done on it. It is hard to tell where you've made a mistake unless you do. You mention $P(x)$ and $r_1$, but where do the $a_k$ come from? $\endgroup$ – robjohn Apr 3 '18 at 17:45
  • $\begingroup$ Write $p(x)=(x-r_1)q(x)+r(x)$, where $q,r$ are polynomials, and $deg(r)<deg(x-r_1)=1$. Therefore, $r$ is a constant polynomial. Since $r_1$ is a root, then $p(r_1)=0$. Evaluating at $x=r_1$ you get $0=p(r_1)=(r_1-r_1)q(r_1)+r=r$. $\endgroup$ – user545963 Apr 3 '18 at 17:49

An easy claim you can prove is:

Claim: When we divide any polynomial $\;P(x)\;$ by a linear polynomial of the form $\;x-r\;$ , the remainder is $\;P(r)\;$ .

And thus if $\;r\;$ is a root of $\;P(x)\;$ the remainder of the wanted division is $\;P(r)=0\;$ .


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.