SOLUTION: Factoring a perfect square trinomial x^3y+2x^2y^2+xy^3

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Question 370325: Factoring a perfect square trinomial
x^3y+2x^2y^2+xy^3

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3 Start with the given expression.


xy%28x%5E2%2B2xy%2By%5E2%29 Factor out the GCF xy.


Now let's try to factor the inner expression x%5E2%2B2xy%2By%5E2


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Looking at the expression x%5E2%2B2xy%2By%5E2, 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 %281%29%281%29=1.


Now the question is: what two whole numbers multiply to 1 (the previous product) and add to the second coefficient 2?


To find these two numbers, we need to list all 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:


First NumberSecond NumberSum
111+1=2
-1-1-1+(-1)=-2



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 and add to 2


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


x%5E2%2Bhighlight%28xy%2Bxy%29%2By%5E2 Replace the second term 2xy with xy%2Bxy.


%28x%5E2%2Bxy%29%2B%28xy%2By%5E2%29 Group the terms into two pairs.


x%28x%2By%29%2B%28xy%2By%5E2%29 Factor out the GCF x from the first group.


x%28x%2By%29%2By%28x%2By%29 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.


%28x%2By%29%28x%2By%29 Combine like terms. Or factor out the common term x%2By


%28x%2By%29%5E2 Condense the terms.


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So xy%28x%5E2%2B2xy%2By%5E2%29 then factors further to xy%28x%2By%29%5E2


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Answer:


So x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3 completely factors to xy%28x%2By%29%5E2.


In other words, x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3=xy%28x%2By%29%5E2.


Note: you can check the answer by expanding xy%28x%2By%29%5E2 to get x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3 or by graphing the original expression and the answer (the two graphs should be identical).


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim