Question 127254
{{{x^2+(x-y)^2-2x*(x-y)}}} Start with the given expression



{{{(x-y)^2-2x*(x-y)+x^2}}} Rearrange the terms



Now let {{{z=x-y}}}. So our equation becomes 



{{{z^2-2xz+x^2}}} Notice how I replaced each {{{x-y}}} with z



Looking at {{{1z^2-2xz+1x^2}}} we can see that the first term is {{{1z^2}}} and the last term is {{{1x^2}}} where the coefficients are 1 and 1 respectively.


Now multiply the first coefficient 1 and the last coefficient 1 to get 1. Now what two numbers multiply to 1 and add to the  middle coefficient -2? Let's list all of the factors of 1:




Factors of 1:

1


-1 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 1

1*1

(-1)*(-1)


note: remember two negative numbers multiplied together make a positive number



Now which of these pairs add to -2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -2


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">1</td><td>1+1=2</td></tr><tr><td align="center">-1</td><td align="center">-1</td><td>-1+(-1)=-2</td></tr></table>



From this list we can see that -1 and -1 add up to -2 and multiply to 1



Now looking at the expression {{{1z^2-2xz+1x^2}}}, replace {{{-2xz}}} with {{{-1xz+-1xz}}} (notice {{{-1xz+-1xz}}} adds up to {{{-2xz}}}. So it is equivalent to {{{-2xz}}})


{{{1z^2+highlight(-1xz+-1xz)+1x^2}}}



Now let's factor {{{1z^2-1xz-1xz+1x^2}}} by grouping:



{{{(1z^2-1xz)+(-1xz+1x^2)}}} Group like terms



{{{z(z-x)-x(z-x)}}} Factor out the GCF of {{{z}}} out of the first group. Factor out the GCF of {{{-x}}} out of the second group



{{{(z-x)(z-x)}}} Since we have a common term of {{{z-x}}}, we can combine like terms


So {{{1z^2-1xz-1xz+1x^2}}} factors to {{{(z-x)(z-x)}}}



So this also means that {{{1z^2-2xz+1x^2}}} factors to {{{(z-x)(z-x)}}} (since {{{1z^2-2xz+1x^2}}} is equivalent to {{{1z^2-1xz-1xz+1x^2}}})



note:  {{{(z-x)(z-x)}}} is equivalent to  {{{(z-x)^2}}} since the term {{{z-x}}} occurs twice. So {{{1z^2-2xz+1x^2}}} also factors to {{{(z-x)^2}}}




So {{{z^2-2xz+1x^2}}} factors to {{{(z-x)^2}}}




{{{(x-y-x)^2}}} Remember, we let {{{z=x-y}}}. So replace each z with {{{x-y}}}



{{{(-y)^2}}} Combine like terms



{{{y^2}}} Square -y to get {{{y^2}}}




------------------------------------


Answer: 



So {{{x^2+(x-y)^2-2x*(x-y)}}} factors to {{{y^2}}}