```Question 409296

Looking at {{{y^2+2xy+x^2}}} we can see that the first term is {{{y^2}}} and the last term is {{{x^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 {{{y^2+2xy+x^2}}}, replace {{{2xy}}} with {{{xy+xy}}} (notice {{{xy+xy}}} adds up to {{{2xy}}}. So it is equivalent to {{{2xy}}})

{{{y^2+highlight(xy+xy)+x^2}}}

Now let's factor {{{y^2+xy+xy+x^2}}} by grouping:

{{{(y^2+xy)+(xy+x^2)}}} Group like terms

{{{y(y+x)+x(y+x)}}} Factor out the GCF of {{{y}}} out of the first group. Factor out the GCF of {{{x}}} out of the second group

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

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

So this also means that {{{y^2+2xy+x^2}}} factors to {{{(y+x)(y+x)}}} (since {{{y^2+2xy+x^2}}} is equivalent to {{{y^2+xy+xy+x^2}}})

note:  {{{(y+x)(y+x)}}} is equivalent to  {{{(y+x)^2}}} since the term {{{y+x}}} occurs twice. So {{{y^2+2xy+x^2}}} also factors to {{{(y+x)^2}}}

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