SOLUTION: Factor the following polynomial completely. 20x2 + 22xy + 6y2 =

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Question 202197: Factor the following polynomial completely.
20x2 + 22xy + 6y2 =
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!


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20x%5E2%2B22xy%2B6y%5E2 Start with the given expression.


2%2810x%5E2%2B11xy%2B3y%5E2%29 Factor out the GCF 2.


Now let's try to factor the inner expression 10x%5E2%2B11xy%2B3y%5E2


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Looking at the expression 10x%5E2%2B11xy%2B3y%5E2, we can see that the first coefficient is 10, the second coefficient is 11, and the last coefficient is 3.


Now multiply the first coefficient 10 by the last coefficient 3 to get %2810%29%283%29=30.


Now the question is: what two whole numbers multiply to 30 (the previous product) and add to the second coefficient 11?


To find these two numbers, we need to list all of the factors of 30 (the previous product).


Factors of 30:
1,2,3,5,6,10,15,30
-1,-2,-3,-5,-6,-10,-15,-30


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


These factors pair up and multiply to 30.
1*30 = 30
2*15 = 30
3*10 = 30
5*6 = 30
(-1)*(-30) = 30
(-2)*(-15) = 30
(-3)*(-10) = 30
(-5)*(-6) = 30

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 11:


First NumberSecond NumberSum
1301+30=31
2152+15=17
3103+10=13
565+6=11
-1-30-1+(-30)=-31
-2-15-2+(-15)=-17
-3-10-3+(-10)=-13
-5-6-5+(-6)=-11



From the table, we can see that the two numbers 5 and 6 add to 11 (the middle coefficient).


So the two numbers 5 and 6 both multiply to 30 and add to 11


Now replace the middle term 11xy with 5xy%2B6xy. Remember, 5 and 6 add to 11. So this shows us that 5xy%2B6xy=11xy.


10x%5E2%2Bhighlight%285xy%2B6xy%29%2B3y%5E2 Replace the second term 11xy with 5xy%2B6xy.


%2810x%5E2%2B5xy%29%2B%286xy%2B3y%5E2%29 Group the terms into two pairs.


5x%282x%2By%29%2B%286xy%2B3y%5E2%29 Factor out the GCF 5x from the first group.


5x%282x%2By%29%2B3y%282x%2By%29 Factor out 3y 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.


%285x%2B3y%29%282x%2By%29 Combine like terms. Or factor out the common term 2x%2By


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So 2%2810x%5E2%2B11xy%2B3y%5E2%29 then factors further to 2%285x%2B3y%29%282x%2By%29


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Answer:


So 20x%5E2%2B22xy%2B6y%5E2 completely factors to 2%285x%2B3y%29%282x%2By%29.


In other words, 20x%5E2%2B22xy%2B6y%5E2=2%285x%2B3y%29%282x%2By%29.


Note: you can check the answer by expanding 2%285x%2B3y%29%282x%2By%29 to get 20x%5E2%2B22xy%2B6y%5E2 or by graphing the original expression and the answer (the two graphs should be identical).

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