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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 $0$?

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  • $\begingroup$ No why would we? Since $x=r$ is a root, $(x-r)$ is a factor! $\endgroup$ – King Tut Apr 3 '18 at 17:43
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    $\begingroup$ 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
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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 $\;P(r)=0\;$ .

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