I have a small problem with long division when dividing polynomials of the same degree I was doing long division with $x^2 + 1$, and $3x^2+5$. (the second polynomial is the quotient).
the problem I've found is related to dividing two polynomials of the same degree. Even if I know that the quotient is always a constant and the remainder is a polynomial of one degree less than the dividend, I still have problem when performing this long division, meaning:
$x^2+0+1\space /\space 3x^2+5$
x^2 is contained in 3x^2 3 times, so I write 3 in the quotient.
then, I perform multiplication between the quotient and the divisor. $3 * 3x^2 = 9x^2$, and $3*5 = 15$. I write them below the dividend.
Now, I subtract the dividend with the things I have below.
but, it's an infinite loop, because the degree doesn't change no matter how long I divide.
 A: You are confusing what you dividing into with what you are dividing by and you are taking the quotient, $3$, and multiply it by what you are dividing into; not what you are dividing by.  You must multiply the quotient by what you are dividing by.
Question 1:  $\frac {3x^2+ 5}{x^2 + 1}$ then we divide $x^2$ into $3x^2$ and get a quotient of $3$.  So we multiply the denominator, $x^2 + 1$ by $3$ to get $3(x^2+1)=3x^2 + 3$.  Then you subtract $(3x^2 + 5)-(3x^2 + 3) = 2$.  Now you have the remainder.
So $\frac {3x^2 + 5}{x^2 + 1} = 3 + \frac 2{x^2 + 1}$.
Question 2: $\frac {x^2 + 1}{3x^2 + 5}$ then we divide $3x^2$ into $x^2$ and we get a quotient of $\frac 13$ (because $3x^2$ goes into $x^2$ a total of $\frac 13$ times).  So we multiply the denominator, $3x^3 + 5$ by $\frac 13$ to got $\frac 13(3x^2 + 5) = x^2 + \frac 53$.  Then we subtract $(x^2 + 1)-(x^2 + \frac 53)= -\frac 23$.  Now we have a remainedr of $-\frac 23$.
So $\frac {x^2+1}{3x^2 + 5} = \frac 13 -\frac {\frac 23}{3x^2 + 5}$.
