<s>[INST] If $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$ is divided by $(x^2 +1)$, then find the remainder 
If the polynomial $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$
is divided by $(x^ 
2
 +1)$, then the remainder is:

How Do I solve this question without the tedious long division?
Using remainder theorem , we can take $x^3$ common and put $x^2 =-1$ although $x$ is not a real number. By this method, I got the right answer as $-x$. Is it the right way? Because $x$ comes out to be $i$ which is not real.
Also , can I apply remainder theorem to quadratic divisor polynomials in this way?
 [/INST] Following may help
$P(x)=x^{19} +x^{17} +x^{13} +x^{11} +x^{7} +x^{5} +x^{3}$
$P(x)=x^{17}(x^2+1)+x^{11}(x^2+1)+x^{5}(x^2+1)+x^3$
$P(x)=(x^2+1)(x^{17}+x^{11}+x^{5})+x^3$
That $x^3$ looks sad alone , Let's also include it
$P(x)=(x^2+1)(x^{17}+x^{11}+x^{5})+x^3+x-x$
$P(x)=(x^2+1)(x^{17}+x^{11}+x^{5}+x)-x$
$P(x)=(x^2+1)(x^{17}+x^{11}+x^{5}+x)-x$
Now this is in the form -
$P(x)=Q(x)*d(x)+R(x)$
Hence you get everything now
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