Question

x^2-2xy+y^2-z^2 factorise this​

Answer 1

Alright... so if there are 4 numbers/variables.. when you take out common factors you have to actually take out common factors. For your answer... (y+z)(y+z) is not taking out a common factor. For y^2-z^2... the common factor is 1.

So since there really are not a lot of common factors in the equation.. look to make a trinomial. For example, seperate the x^2+2xy+y^2. This results in a perfect square. A perfect square a variable + another squared. In this case... the perfect square of x^2+2xy+y^2 is (x+z)(x+z) or (x+z)^2. Therefore... you are left with (x+z)(x+z)-z^2

Answer 2

Answer:

$\huge\bold\green{Given:}$

$\huge\bold{{x}^2\:-\:2(xy)\:+\:{y}^2\:-\:{z}^2}$

$\huge\bold\green{Solution:}$

$\huge\bold{{x}^2\:-\:2(xy)\:+\:{y}^2\:-\:{z}^2}$

By using : $\huge\bold(a\:-\:b)^2\:=\:a^2\:+\:b^2\:-\:2(ab)$

$\huge\bold{(x\:-\:y)^2\:-\:{z}^2}$

$\huge\bold{(x\:-\:y\:+\:z)(x\:-\:y\:-\:z)}$