Need help on dividing polynomials I have trouble understanding how to divide  $x^4 + y^4$ by $f_1 = x^2 + y$ and $f_2 =  x^2 y + 1$ using the ordering $ y \leq x$ and separately for $ x \leq y$. 
Please help! I went to the tutoring center but none of the specialists understand this and I have never done this before in high school.
 A: Note: I am making an assumption here about exactly what is wanted, but I think it the likeliest possibility for a pre-calculus algebra course.
For the order $y\le x$, pretend that $y$ is a constant. Then you get this long division:
$$\require{enclose}\begin{align}
x^2-\phantom{-x^2}y\phantom{+y^4+y^2}\\
x^2+y\enclose{longdiv}{x^4+\phantom{-x^2y}+\phantom{y^4+}y^4}\\
\underline{x^4+\phantom{-}x^2y\,\phantom{+y^4+y^2}}\\
-x^2y+\phantom{y^4+}y^4\\
\underline{-x^2y-\,\phantom{y^4+}y^2}\\
y^4+y^2
\end{align}$$
Thus, $x^4+y^4=\left(x^2+y\right)\left(x^2-y\right)+\left(y^4+y^2\right)$, with quotient $x^2-y$ and remainder $y^4+y^2$.
For the order $x\le y$ you treat $x$ as if it were a constant:
$$\require{enclose}\begin{align}
y^3-\phantom-x^2y^2+x^4y-\phantom{-y}x^6\;\phantom{+x^8+x^4}\\
y+x^2\enclose{longdiv}{y^4+\phantom{-x^2y^3-x^4y^2+-x^6y}+\phantom{x^8+}x^4}\\
\underline{y^4+\phantom-x^2y^3\;\phantom{-x^4y^2+-x^6y+x^8+x^4}}\\
-x^2y^3\,\phantom{-x^4y^2+-x^6y+x^8+x^4}\\
\underline{-x^2y^3-x^4y^2\phantom{+-x^6y+x^8+x^4}}\\
x^4y^2\phantom{+-x^6y+x^8+x^4}\\
\underline{x^4y^2+\phantom-x^6y\phantom{+x^8+x^4}\;\;}\\
-x^6y+\;\phantom{x^8+}x^4\\
\underline{-x^6y-\;\phantom{x^8+}x^8}\\
x^8+x^4
\end{align}$$
That is, $y^4+x^4=\left(y+x^2\right)\left(y^3-x^2y^2+x^4y-x^6\right)+\left(x^8+x^4\right)$, with quotient $$y^3-x^2y^2+x^4y-x^6$$ and remainder $x^8+x^4$.
