SOLUTION: Can someone walk me through the steps for the following problem? Thank you! x^2-3xy+9y^2 1. Factor completely. 2. Remember to look first for a common factor. 3. Check by

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Can someone walk me through the steps for the following problem? Thank you! x^2-3xy+9y^2 1. Factor completely. 2. Remember to look first for a common factor. 3. Check by      Log On


   

Question 674305: Can someone walk me through the steps for the following problem? Thank you!
x^2-3xy+9y^2
1. Factor completely.
2. Remember to look first for a common factor.
3. Check by multiplying.
4. If a polynomial is prime, state this.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression x%5E2-3xy%2B9y%5E2, we can see that the first coefficient is 1, the second coefficient is -3, and the last coefficient is 9.


Now multiply the first coefficient 1 by the last coefficient 9 to get %281%29%289%29=9.


Now the question is: what two whole numbers multiply to 9 (the previous product) and add to the second coefficient -3?


To find these two numbers, we need to list all of the factors of 9 (the previous product).


Factors of 9:
1,3,9
-1,-3,-9


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


These factors pair up and multiply to 9.
1*9 = 9
3*3 = 9
(-1)*(-9) = 9
(-3)*(-3) = 9

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -3:


First NumberSecond NumberSum
191+9=10
333+3=6
-1-9-1+(-9)=-10
-3-3-3+(-3)=-6



From the table, we can see that there are no pairs of numbers which add to -3. So x%5E2-3xy%2B9y%5E2 cannot be factored.


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Answer:


So x%5E2-3xy%2B9y%5E2 doesn't factor at all (over the rational numbers).


So x%5E2-3xy%2B9y%5E2 is prime.