x^2-y^2+2zx+2yz+2y-2y-1
To solve it you’re supposed to arrange the expression in standard polynomial form, although from there I get a bit lost.
The answer looks like...
(x+y-1)(x-y+2z+1)
Looking at your expression,
x^2 – y^2 + 2zx + 2yz + 2y – 2y – 1
Now at the end there you have 2y – 2y – 1, which doesn’t make much sense since 2y – 2y = 0!
So I’m guessing that this is a typo and that it is supposed to be 2y – 2z – 1. This gives us for the original expression,
x^2 – y^2 + 2zx + 2yz + 2y – 2z – 1
A second assumption of mine here is that you have been given the final answer, (x+y-1)(x-y+2z+1), as part of the problem.
We can write the new original expression as,
x^2 – y^2 + z(2x + 2y – 2) + 2y – 1
x^2 – y^2 + 2y – 1 + 2z(x + y – 1)
Since we are given the final answer, (x + y – 1)(x – y + 2z + 1), which has a common factor of (x + y – 1), and since we already have one instance of this factor with 2z(x + y – 1), then we now need to show that (x + y – 1) is also a factor of the remaining term, x^2 – y^2 + 2y – 1.
(x + y – 1) ) x^2 – y^2 + 2y – 1 ( x – y + 1
x^2 + xy – x
===============
-xy – y^2 + x + 2y – 1
-xy – y^2 + y
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x + y – 1
x + y – 1
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0
So, for x^2 – y^2 + 2y – 1, we can substitute (x + y – 1)( x – y + 1)
Our expression now is,
(x + y – 1)( x – y + 1) + 2z(x + y – 1)
Taking out the common factor (x + y – 1) gives us,
(x + y – 1)( x – y + 1 + 2z)