Factorization help please!
original equation:xsquared-ysquared + 2zx +2yz+27 -2z -1
The furthest step i got to: xsquared + (2z)x -[(y-1){y-(2z + 1)]
What do you do now?

Question

mr.math · Answer

There are more than one factorization for such an expression.

mr.math · Answer

$$x^2-y^2+2zx +2yz+27 -2z -1=(x+y)(x-y)+2z(x+y)+26-2z$$
$$=(x-y)(x+y+2z)+2(12-z)$$

anonymous · Answer

(x+y)(x+2z-y)-2z+26

mr.math · Answer

Oh, I've got a better factorization.

anonymous · Answer

Mr.Math its 13-z at last i think

anonymous · Answer

I messed up the furthest step I got to
I got :
x squared+ (2z)x- [y squared - (2z+2)y +(2z+1) ]
They just want factorization, no 'answers'

mr.math · Answer

Yeah, 13-z.

anonymous · Answer

$$(x+y)(x+2z-y)-2(z-13)$$

anonymous · Answer

And it is x+y which has to be taken common over there
See the above post I gave the answer to your thought

anonymous · Answer

oh I got something completely different
(x+y-1) (x-y+2z+1)

anonymous · Answer

Hopex its perfect dude U can go for it

mr.math · Answer

I don't think that's right. @hopex

anonymous · Answer

ohhh sry sry wait wait u will get something to be added at last some constant

anonymous · Answer

oh wait the original equation is actually :
xsquared - y squared + 2zx+2yz+2y-2z-1
 Messed that up sorry!

anonymous · Answer

ohhh dude see that first buddy i thought its z at first but I thought it is correct as given by u

anonymous · Answer

So I got to the step xsquared +(2z)x -[ysquared - (2z +2)y + (2z+1)]
but how does that factor into xsquared +(2z)x - (y -1) [y-(2z +1)]
I don't get that step^