SOLUTION: help me solve and factor x^2+12xy+32y^2

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Question 645121: help me solve and factor x^2+12xy+32y^2
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression x%5E2%2B12xy%2B32y%5E2, we can see that the first coefficient is 1, the second coefficient is 12, and the last coefficient is 32.


Now multiply the first coefficient 1 by the last coefficient 32 to get %281%29%2832%29=32.


Now the question is: what two whole numbers multiply to 32 (the previous product) and add to the second coefficient 12?


To find these two numbers, we need to list all of the factors of 32 (the previous product).


Factors of 32:
1,2,4,8,16,32
-1,-2,-4,-8,-16,-32


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


These factors pair up and multiply to 32.
1*32 = 32
2*16 = 32
4*8 = 32
(-1)*(-32) = 32
(-2)*(-16) = 32
(-4)*(-8) = 32

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 12:


First NumberSecond NumberSum
1321+32=33
2162+16=18
484+8=12
-1-32-1+(-32)=-33
-2-16-2+(-16)=-18
-4-8-4+(-8)=-12



From the table, we can see that the two numbers 4 and 8 add to 12 (the middle coefficient).


So the two numbers 4 and 8 both multiply to 32 and add to 12


Now replace the middle term 12xy with 4xy%2B8xy. Remember, 4 and 8 add to 12. So this shows us that 4xy%2B8xy=12xy.


x%5E2%2Bhighlight%284xy%2B8xy%29%2B32y%5E2 Replace the second term 12xy with 4xy%2B8xy.


%28x%5E2%2B4xy%29%2B%288xy%2B32y%5E2%29 Group the terms into two pairs.


x%28x%2B4y%29%2B%288xy%2B32y%5E2%29 Factor out the GCF x from the first group.


x%28x%2B4y%29%2B8y%28x%2B4y%29 Factor out 8y 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%2B8y%29%28x%2B4y%29 Combine like terms. Or factor out the common term x%2B4y


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Answer:


So x%5E2%2B12xy%2B32y%5E2 factors to %28x%2B8y%29%28x%2B4y%29.


In other words, x%5E2%2B12xy%2B32y%5E2=%28x%2B8y%29%28x%2B4y%29.


Note: you can check the answer by expanding %28x%2B8y%29%28x%2B4y%29 to get x%5E2%2B12xy%2B32y%5E2 or by graphing the original expression and the answer (the two graphs should be identical).

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