SOLUTION: I am having trouble factoring the following equation: 2a^2-9a+3

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I am having trouble factoring the following equation: 2a^2-9a+3      Log On


   

Question 848184: I am having trouble factoring the following equation: 2a^2-9a+3
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
It isn't an equation. Do you see an equal sign?
It is a polynomial expression.
It can't be factored.
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 2a%5E2-9a%2B3, we can see that the first coefficient is 2, the second coefficient is -9, and the last term is 3.



Now multiply the first coefficient 2 by the last term 3 to get %282%29%283%29=6.



Now the question is: what two whole numbers multiply to 6 (the previous product) and add to the second coefficient -9?



To find these two numbers, we need to list all of the factors of 6 (the previous product).



Factors of 6:

1,2,3,6

-1,-2,-3,-6



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



These factors pair up and multiply to 6.

1*6 = 6
2*3 = 6
(-1)*(-6) = 6
(-2)*(-3) = 6


Now let's add up each pair of factors to see if one pair adds to the middle coefficient -9:



First NumberSecond NumberSum
161+6=7
232+3=5
-1-6-1+(-6)=-7
-2-3-2+(-3)=-5




From the table, we can see that there are no pairs of numbers which add to -9. So 2a%5E2-9a%2B3 cannot be factored.



===============================================================





Answer:



So 2%2Aa%5E2-9%2Aa%2B3 doesn't factor at all (over the rational numbers).



So 2%2Aa%5E2-9%2Aa%2B3 is prime.