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

Question

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?

Answer 1

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