SOLUTION: Factor completely a^2-2ab+5

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Question 751412: Factor completely
a^2-2ab+5
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression a%5E2-2ab%2B5, we can see that the first coefficient is 1, the second coefficient is -2, and the last coefficient is 5.


Now multiply the first coefficient 1 by the last coefficient 5 to get %281%29%285%29=5.


Now the question is: what two whole numbers multiply to 5 (the previous product) and add to the second coefficient -2?


To find these two numbers, we need to list all of the factors of 5 (the previous product).


Factors of 5:
1,5
-1,-5


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


These factors pair up and multiply to 5.
1*5 = 5
(-1)*(-5) = 5

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -2:


First NumberSecond NumberSum
151+5=6
-1-5-1+(-5)=-6



From the table, we can see that there are no pairs of numbers which add to -2. So a%5E2-2ab%2B5 cannot be factored.


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Answer:


So a%5E2-2ab%2B5 doesn't factor at all (over the rational numbers).


So a%5E2-2ab%2B5 is prime.


Note: there may be a typo, so make sure everything is typed correctly, thanks.