X^2-y^2 = (x+y)(x-y)

  • 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?
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
132


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?
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.

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?
 

Related Threads on X^2-y^2 = (x+y)(x-y)

  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
10K
  • Last Post
Replies
5
Views
11K
  • Last Post
Replies
2
Views
2K
Replies
6
Views
3K
  • Last Post
Replies
2
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
3
Views
549
Replies
7
Views
2K
Top