# can someone please factor this???

• Mar 3rd 2009, 11:02 PM
bcguy87
can someone please factor this for me?? if it cant be factored can you explain why? thanks

(x-y)(x^2-z^2)-(x-z)(x^2-y^2)
• Mar 3rd 2009, 11:23 PM
earboth
Quote:

Originally Posted by bcguy87
can someone please factor this for me?? if it cant be factored can you explain why? thanks

(x-y)(x^2-z^2)-(x-z)(x^2-y^2)

You are supposed to know that

$\displaystyle a^2-b^2=(a+b)(a-b)$

Therefore you get:

$\displaystyle (x-y)(x^2-z^2)-(x-z)(x^2-y^2) = (x-y) \left((x+z)(x-z) \right) - (x-z)\left((x+y)(x-y) \right)$

You see that the common factor is (x-y)(x-z). Factor out:

$\displaystyle (x-y) \left((x+z)(x-z) \right) - (x-z)\left((x+y)(x-y) \right) =$ $\displaystyle (x-y)(x-z) \left((x+z) - (x+y) \right) = (x-y)(x-z)(z-y)$
• Mar 3rd 2009, 11:31 PM
bcguy87
ok, so how come when u work backwards now from what you got, the answer is not anywhere near what the question was??
because, when u reverse your factoring method, you are supposed to get the original question................

i took your $\displaystyle a^2-b^2$ then i worked it out to get $\displaystyle a^2-ab+ab-b^2$ then when worked back i got your original question..............just a thought
• Mar 4th 2009, 04:13 AM
earboth
Quote:

Originally Posted by bcguy87
ok, so how come when u work backwards now from what you got, the answer is not anywhere near what the question was??
...

Take your original term and expand the brackets:

$\displaystyle (x-y)(x^2-z^2)-(x-z)(x^2-y^2) = x^2·z - x^2·y - x·z^2 + x·y^2 + y·z^2 - y^2·z$

Now take my result and expand the brackets:

$\displaystyle (x-y)(x-z)(z-y) = - x^2·y + x^2·z + x·y^2 - x·z^2 - y^2·z + y·z^2$

If you compare both results you'll see that they are equal.