Question 190364


{{{x^3y+2x^2y^2+xy^3}}} Start with the given expression



{{{xy(x^2+2xy+y^2)}}} Factor out the GCF {{{xy}}}



Now let's focus on the inner expression {{{x^2+2xy+y^2}}}





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


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



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



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



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



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


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



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



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




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So our expression goes from {{{xy(x^2+2xy+y^2)}}} and factors further to {{{xy(x+y)^2}}}



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Answer:


So {{{x^3y+2x^2y^2+xy^3}}} factors to {{{xy(x+y)^2}}}