# How do you divide 2x^4 - x^3y + x^2y^2 + 4xy^3 - 4y^4 div 2x^2 + xy - 2y^2?

Jun 27, 2016

${x}^{2} - x y + 2 {y}^{2}$

#### Explanation:

Calling

$N \left(x , y\right) = 2 {x}^{4} - {x}^{3} y + {x}^{2} {y}^{2} + 4 x {y}^{3} - 4 {y}^{4}$
$D \left(x , y\right) = 2 {x}^{2} + x y - 2 {y}^{2}$

Dividing $N \left(x , y\right)$ into $D \left(x , y\right)$ will give a result $Q \left(x , y\right)$ with maximum $x$ coefficient equal $2$ and maximum $y$ coefficient equal $2$.
The remainder $R \left(x , y\right)$ will have maximum $x$ coefficient equal $1$ and maximum $y$ coefficient equal $1$. So

$Q \left(x , y\right) = {\sum}_{i = 0 , j = 0}^{i = 2 , j = 2} {q}_{i j} {x}^{i} {y}^{j}$

and

$R \left(x , y\right) = {r}_{11} x y + {r}_{10} x + {r}_{01} y + {r}_{00}$

Expanding

$N \left(x , y\right) = D \left(x , y\right) Q \left(x , y\right) + R \left(x , y\right)$

and equating the coefficients we have (I am considering only nontrivial relationships)

{ ( -4 + 2 q_{02}=0), (-q_{00} - r_{11}=0),( -q_{01} + 2 q_{10}=0), (4 - q_{02} + 2 q_{11}=0), (-2 q_{01} - q_{10}=0), (1 - 2 q_{02} - q_{11} + 2 q_{20}=0), (-q_{12} + 2 q_{21}=0), (-1 - 2 q_{11} - q_{20}=0), (-2 q_{12} - q_{21}=0), (2 - 2 q_{20}=0) :}

solving we have

${q}_{02} = 1 , {q}_{11} = - 1 , {q}_{20} = 1$ and all others nulls. So the result is

$\frac{N \left(x , y\right)}{D \left(x , y\right)} = {x}^{2} - x y + 2 {y}^{2}$