SOLUTION: How do you factor x^2 + 5xy - 14y^2 ?

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Question 599427: How do you factor x^2 + 5xy - 14y^2 ?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression x%5E2%2B5xy-14y%5E2, we can see that the first coefficient is 1, the second coefficient is 5, and the last coefficient is -14.


Now multiply the first coefficient 1 by the last coefficient -14 to get %281%29%28-14%29=-14.


Now the question is: what two whole numbers multiply to -14 (the previous product) and add to the second coefficient 5?


To find these two numbers, we need to list all of the factors of -14 (the previous product).


Factors of -14:
1,2,7,14
-1,-2,-7,-14


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


These factors pair up and multiply to -14.
1*(-14) = -14
2*(-7) = -14
(-1)*(14) = -14
(-2)*(7) = -14

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 5:


First NumberSecond NumberSum
1-141+(-14)=-13
2-72+(-7)=-5
-114-1+14=13
-27-2+7=5



From the table, we can see that the two numbers -2 and 7 add to 5 (the middle coefficient).


So the two numbers -2 and 7 both multiply to -14 and add to 5


Now replace the middle term 5xy with -2xy%2B7xy. Remember, -2 and 7 add to 5. So this shows us that -2xy%2B7xy=5xy.


x%5E2%2Bhighlight%28-2xy%2B7xy%29-14y%5E2 Replace the second term 5xy with -2xy%2B7xy.


%28x%5E2-2xy%29%2B%287xy-14y%5E2%29 Group the terms into two pairs.


x%28x-2y%29%2B%287xy-14y%5E2%29 Factor out the GCF x from the first group.


x%28x-2y%29%2B7y%28x-2y%29 Factor out 7y 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%2B7y%29%28x-2y%29 Combine like terms. Or factor out the common term x-2y


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Answer:


So x%5E2%2B5xy-14y%5E2 factors to %28x%2B7y%29%28x-2y%29.


In other words, x%5E2%2B5xy-14y%5E2=%28x%2B7y%29%28x-2y%29.


Note: you can check the answer by expanding %28x%2B7y%29%28x-2y%29 to get x%5E2%2B5xy-14y%5E2 or by graphing the original expression and the answer (the two graphs should be identical).