Question 370325
{{{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 try to factor the inner expression {{{x^2+2xy+y^2}}}



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Looking at the expression {{{x^2+2xy+y^2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{2}}}, and the last coefficient is {{{1}}}.



Now multiply the first coefficient {{{1}}} by the last coefficient {{{1}}} to get {{{(1)(1)=1}}}.



Now the question is: what two whole numbers multiply to {{{1}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{2}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{1}}} (the previous product).



Factors of {{{1}}}:

1

-1



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{1}}}.

1*1 = 1
(-1)*(-1) = 1


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{2}}}:



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



From the table, we can see that the two numbers {{{1}}} and {{{1}}} add to {{{2}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{1}}} both multiply to {{{1}}} <font size=4><b>and</b></font> add to {{{2}}}



Now replace the middle term {{{2xy}}} with {{{xy+xy}}}. Remember, {{{1}}} and {{{1}}} add to {{{2}}}. So this shows us that {{{xy+xy=2xy}}}.



{{{x^2+highlight(xy+xy)+y^2}}} Replace the second term {{{2xy}}} with {{{xy+xy}}}.



{{{(x^2+xy)+(xy+y^2)}}} Group the terms into two pairs.



{{{x(x+y)+(xy+y^2)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+y)+y(x+y)}}} Factor out {{{y}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(x+y)(x+y)}}} Combine like terms. Or factor out the common term {{{x+y}}}



{{{(x+y)^2}}} Condense the terms.



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



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Answer:



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



In other words, {{{x^3y+2x^2y^2+xy^3=xy(x+y)^2}}}.



Note: you can check the answer by expanding {{{xy(x+y)^2}}} to get {{{x^3y+2x^2y^2+xy^3}}} or by graphing the original expression and the answer (the two graphs should be identical).



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Jim