- #1

- 95

- 0

So

[tex]x^2-y^2=(x+y)(x-y)[/tex]

in the same sense what does

[tex]x^2-y^2-z^2=?[/tex]

come to?

[tex]x^2-y^2=(x+y)(x-y)[/tex]

in the same sense what does

[tex]x^2-y^2-z^2=?[/tex]

come to?

- Thread starter JDude13
- Start date

- #1

- 95

- 0

So

[tex]x^2-y^2=(x+y)(x-y)[/tex]

in the same sense what does

[tex]x^2-y^2-z^2=?[/tex]

come to?

[tex]x^2-y^2=(x+y)(x-y)[/tex]

in the same sense what does

[tex]x^2-y^2-z^2=?[/tex]

come to?

- #2

chiro

Science Advisor

- 4,790

- 132

You are not going to necessarily get the kind of factorization you got with two terms. If your x^2 - y^2 was itself a positive number, you could apply the same formula that you used for x and y.So

[tex]x^2-y^2=(x+y)(x-y)[/tex]

in the same sense what does

[tex]x^2-y^2-z^2=?[/tex]

come to?

If however your x^2 - y^2 was negative you would get a negative term - a negative term which would be in the form -(a + b^2) (a, b^2 >= 0) which has no standard factorization.

- #3

- 95

- 0

Ive had a muck around with it and

[tex]x^2-y^2-z^2=(x+y+z)(x-y-z)+yz[/tex]

- #4

- 5,439

- 9

shouldn't that be +2yz?

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