I'm having problems dividing the polynomial 2x^5 + 2x^3 + x^2 + 1 by x^2 + 1
Looks straight forward to me. But I don't know what to tell you without seeing what you did. Did you remember to include "0x" in the divisor and "$\displaystyle 0x^4$" and "$\displaystyle 0x$" in the dividend? How many times does $\displaystyle x^2$ divide into $\displaystyle 2x^5$?
$\displaystyle x^2+0x+1sqrt(2x^5+0x^4+2x^3+x^2+0x+1)$
This is the subtracting parts: The quotient i got is $\displaystyle 2x^3+x^2+x+1$
Part 1 $\displaystyle 2x^5+0x^4+2x^3$
Part 2 $\displaystyle x^4+x^2$
$\displaystyle x^4+x^3+x^2$
Part 3 $\displaystyle x^3+0x$
$\displaystyle x^3+x^2+x$
Part 4 $\displaystyle x^2+1$
$\displaystyle x^2+x+1$
Final $\displaystyle x$
Yes, $\displaystyle x^2$ divides into $\displaystyle 2x^5$ $\displaystyle 2x^3$ times. And multiplying the entire divisor by $\displaystyle 2x^3$ gives $\displaystyle 2x^5+ 2x^3$
Now subtract that from the dividend:
$\displaystyle 2x^5+ 2x^3+ x^2+ 1- (2x^5+ 2x^3)= x^2+ 1$.
And now, of course, $\displaystyle x^2+ 1$ divides into that exactly once: the quotient is $\displaystyle 2x^3+ 1$.