x^2-y^2+2zx+2yz+2y-2y-1

Question

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)

Answer 1

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

                       =================

                                           x + y – 1

                                           x + y – 1

                                          ========

                                                  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)

 

 

 

Answer 2

Using the answer as a guide, we can see the expression should be:

x²-y²+2zx+2yz+2y-2z-1

From now on, we will not use the expected answer as a guide to the solution.

This expression can be rearranged:

x²-y²+2y-1+2z(x+y-1)=x²-(y-1)²+2z(x+y-1)

The first part of this is the difference of two squares which factorises:

(x-(y-1))(x+(y-1))+2z(x+y-1)=(x-y+1)(x+y-1)+2z(x+y-1).

There’s a common factor of x+y-1 which can be taken out:

(x+y-1)(x-y+1+2z) which is effectively the same as the given answer.