<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] With $y:=x^2$,$$P(x)=x^{19} +x^{17} +x^ {13} +x^ {11} +x^ 7 +x^ 5 +x^ 3=(y^9+y^8+y^6+y^5+y^3+y^2+y)x$$ and you are asked the remainder of the division by the polynomial $x^2+1=y+1$.
You can write
$$P(x)=(Q(y)(y+1)+R(y))x$$ where $R$ is of degree $0$, i.e. a constant (which you know how to compute by division by a linear binomial).
From this
$$P(x)=Q(x^2)(x^2+1)x+Rx$$ and the remainder is $Rx$.

Note that at no time do you have to deal with the solutions of $x^2+1=0$.
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