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#### paulmdrdo

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- May 13, 2013

- 386

$\displaystyle \frac{2xz-8xy-8yz+z^2+x^2+16y^2}{x-4y-z}$

and also i'm solving this using the algorithm used in arithmetic division is there a way to type that using latex?

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- Thread starter
- #1

- May 13, 2013

- 386

$\displaystyle \frac{2xz-8xy-8yz+z^2+x^2+16y^2}{x-4y-z}$

and also i'm solving this using the algorithm used in arithmetic division is there a way to type that using latex?

- Feb 13, 2012

- 1,704

Take into account that...

$\displaystyle \frac{2xz-8xy-8yz+z^2+x^2+16y^2}{x-4y-z}$

and also i'm solving this using the algorithm used in arithmetic division is there a way to type that using latex?

$\displaystyle (a\ x + b\ y + c\ z)^{2} = a^{2}\ x^{2} + b^{2}\ y^{2} + c^{2}\ y^{2} + 2\ a\ b\ x\ y\ + 2\ a\ c\ x\ z\ + 2\ b\ c\ y\ z\ (1)$

Kind regards

$\chi$ $\sigma$

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- May 13, 2013

- 386

what if i have something like

$9m^3n-32m^3+42m^6-15n^2-n+6$

I was trying to solve this problem when i got confused on

how to arrange the terms in descending powers of the literal factors

because some term contain two variables and the polynomial has 3 variables.

How can i properly arrange the dividend here?

$\displaystyle \frac{2xz-8xy-8yz+z^2+x^2+16y^2}{x-4y-z}$

It's tricky to explain my procedure . . .

Since the divisor has [tex]x,y,z[/tex], I took the terms with [tex]x^2,\,xy,\,xz[/tex]

. . [tex]x^2 - 8xy + 2xz[/tex]

Then I took the terms with [tex]y^2,\,yz[/tex]

. . [tex]16y^2 - 8yz + z^2[/tex]

So we have: .[tex]\frac{x^2-8xy + 2x z + 16y^2 - 8yz + z^2}{x-4y-z}[/tex]

[tex]\begin{array}{cccccccccccccc} &&&&&& x &&& - & 4y & + & 3z \\ && --&--&--&--&--&--&--&--&--& --&-- \\ x-4y-z & | & x^2 &-& 8xy &+& 2xz &+&16y^2 &-& 8yz &+& z^2 \\ &&x^2 &-& 4xy &-& xz \\ && --&--&--&--&-- \\ &&& -& 4xy &+& 3xz &+& 16y^2 &-& 8yz &+& z^2 \\ &&& - & 4xy &+& & +& 16y^2 &+& 4yz \\ &&& --&--&--&--&--&--&--&--&--&-- \\ &&&&&& 3xz &&& - & 12yz &+& z^2 \\ &&&&&& 3xz &&& -& 12yz &-& 3z^2 \\ &&&&&& --&--&--&--&--&--&-- \\ &&&&&&&&&&&& 4z^2 \end{array}[/tex]

Therefore: ..[tex]\frac{x^2-8xy + 2x z + 16y^2 - 8yz + z^2}{x-4y-z} \;\;=\;\;x - 4y + 3x + \frac{4z^2}{x-4y-z}[/tex]

- Feb 21, 2013

- 739

I had Also hard to understand long polynom division but after watching this video evrything made sense! So I Will recomend him to watch this and it should be easy to understand http://m.youtube.com/watch?v=l6_ghhd7kwQHello, paulmdrdo!

It's tricky to explain my procedure . . .

Since the divisor has [tex]x,y,z[/tex], I took the terms with [tex]x^2,\,xy,\,xz[/tex]

. . [tex]x^2 - 8xy + 2xz[/tex]

Then I took the terms with [tex]y^2,\,yz[/tex]

. . [tex]16y^2 - 8yz + z^2[/tex]

So we have: .[tex]\frac{x^2-8xy + 2x z + 16y^2 - 8yz + z^2}{x-4y-z}[/tex]

[tex]\begin{array}{cccccccccccccc} &&&&&& x &&& - & 4y & + & 3z \\ && --&--&--&--&--&--&--&--&--& --&-- \\ x-4y-z & | & x^2 &-& 8xy &+& 2xz &+&16y^2 &-& 8yz &+& z^2 \\ &&x^2 &-& 4xy &-& xz \\ && --&--&--&--&-- \\ &&& -& 4xy &+& 3xz &+& 16y^2 &-& 8yz &+& z^2 \\ &&& - & 4xy &+& & +& 16y^2 &+& 4yz \\ &&& --&--&--&--&--&--&--&--&--&-- \\ &&&&&& 3xz &&& - & 12yz &+& z^2 \\ &&&&&& 3xz &&& -& 12yz &-& 3z^2 \\ &&&&&& --&--&--&--&--&--&-- \\ &&&&&&&&&&&& 4z^2 \end{array}[/tex]

Therefore: ..[tex]\frac{x^2-8xy + 2x z + 16y^2 - 8yz + z^2}{x-4y-z} \;\;=\;\;x - 4y + 3x + \frac{4z^2}{x-4y-z}[/tex]

Ps. You Really got nice latex skill, if I would do that I would just screw up and give up

Regards,

\(\displaystyle |\pi\rangle\)