Polynomial long division with x and y

Problem: Use polynomial long division to find the extreme point to

$\displaystyle f\left( x,y \right)=\left( x^{2}+y^{2} \right)\left( xy+1 \right)=x^{3}y+x^{2}+xy^{3}+y^{2}$

Attempt:

$\displaystyle f\left( x,y \right)=x^{3}y+x^{2}+xy^{3}+y^{2}$

$\displaystyle \frac{\partial f}{\partial x}=3x^{2}y+2x+y^{3}$

$\displaystyle \frac{\partial f}{\partial y}=x^{3}+3xy^{2}+2y$

I see that (0,0) is an extreme point, but I havn't learned how to use polynomial long division with more than one variable. How do I do it when I have both x and y?

Re: Polynomial long division with x and y

Re: Polynomial long division with x and y

MaxJasper, does that mean 0,0 is the only point? Still: the only tools I'm allowed to have is pen, paper and—of course—polynomial long division.

Re: Polynomial long division with x and y

What do you define extreme point for such f(x,y)?

Re: Polynomial long division with x and y

Defn for extreme: ∂f/∂x=∂f/dy=0.

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Re: Polynomial long division with x and y

Quote:

Originally Posted by

**jacob93** Defn for extreme: ∂f/∂x=∂f/dy=0.

There are nine extreme points :

$\displaystyle \{x,y\}=$

$\displaystyle \{0,0\}$

$\displaystyle \left\{-(-1)^{1/4},-(-1)^{3/4}\right\}$

$\displaystyle \left\{(-1)^{1/4},(-1)^{3/4}\right\}$

$\displaystyle \left\{-(-1)^{3/4},-(-1)^{1/4}\right\}$

$\displaystyle \left\{(-1)^{3/4},(-1)^{1/4}\right\}$

$\displaystyle \left\{-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\}$

$\displaystyle \left\{-\frac{i}{\sqrt{2}},-\frac{i}{\sqrt{2}}\right\}$

$\displaystyle \left\{\frac{i}{\sqrt{2}},\frac{i}{\sqrt{2}}\right \}$

$\displaystyle \left\{\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right\}$

http://mathhelpforum.com/attachment....1&d=1347645744

Re: Polynomial long division with x and y

That's good, although we don't use $\displaystyle i$ in this course. My question is this: how did you get to those points? I want to learn how to find *all* the solution.

Re: Polynomial long division with x and y

Solve equations you specified: ∂f/∂x=∂f/dy=0.

Re: Polynomial long division with x and y

Obviously, but is there a systematic way of doing so?

Re: Polynomial long division with x and y

Hello, jacob93!

Quote:

Use polynomial long division to find the extreme points to

$\displaystyle f\left( x,y \right)\:=\:\left( x^{2}+y^{2} \right)\left( xy+1 \right)=x^{3}y+x^{2}+xy^{3}+y^{2}$

Attempt: $\displaystyle f\left( x,y \right)=x^{3}y+x^{2}+xy^{3}+y^{2}$

$\displaystyle \frac{\partial f}{\partial x}=3x^{2}y+2x+y^{3}$

$\displaystyle \frac{\partial f}{\partial y}=x^{3}+3xy^{2}+2y$

I see that (0,0) is an extreme point, but I havn't learned how to use polynomial long division

with more than one variable. .How do I do it when I have both x and y?

I'm not sure how long division can be applied,

. . but I found some factoring . . . and that's all.

We have: .$\displaystyle \begin{array}{ccccccc}f_x &=& 3x^2y + 2x + y^3 &=& 0 & [1] \\ f_y &=& x^3 + 3xy^2 + 2y & = & 0 &[2] \end{array}$

Subtract [2] - [1]: .$\displaystyle x^3 - y^3- 3x^2y + 3xy^2 - 2x + 2y \:=\:0$

. . . . . $\displaystyle (x-y)(x^2+xy + y^2) - 3xy(x - y) - 2(x - y) \:=\:0$

m . . . . . . . . . . . . . $\displaystyle (x-y)(x^2+xy+y^2-3xy - 2) \:=\:0$

n . . . . . . . . . . . . . . . . . $\displaystyle (x-y)(x^2-2xy + y^2 - 2) \:=\:0$

. . . . . . . . . . . . . . . . . . . . . .$\displaystyle (x-y)\left([x-y]^2-2\right) \:=\:0$

Hence we have: .$\displaystyle \begin{Bmatrix}x-y \:=\:0 & \Rightarrow & y\:=\:x \\ (x-y)^2-2 \:=\:0 & \Rightarrow & x-y \:=\:\pm\sqrt{2} \end{Bmatrix}$

You take it from here . . .

Re: Polynomial long division with x and y

I saw you did [2]-[1] and I tried it myself. Couldn't do it. Looked at the rest of your post and realized I would not have thought of factoring out (x-y) which was essential. Do I lack some intuition or skill?

Re: Polynomial long division with x and y

Tried using the cubic rule now. $\displaystyle \left[ 1 \right]+\left[ 2 \right]\; \Rightarrow\; \left( x+y \right)^{3}+2\left( y+x \right)\; \Rightarrow\; x=-y$