SOLUTION: I can't find the answer to the following question on my assignment... I have to use the window or box method to factor this equation... 4x^2-6x+8 If I could get the factors

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I can't find the answer to the following question on my assignment... I have to use the window or box method to factor this equation... 4x^2-6x+8 If I could get the factors       Log On


   



Question 618245: I can't find the answer to the following question on my assignment... I have to use the window or box method to factor this equation...
4x^2-6x+8
If I could get the factors I know how to do the box but I can't get it factored.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
4x^2-6x+8
factor out 2
2x^2-3x+4
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 2x%5E2-3x%2B4, we can see that the first coefficient is 2, the second coefficient is -3, and the last term is 4.



Now multiply the first coefficient 2 by the last term 4 to get %282%29%284%29=8.



Now the question is: what two whole numbers multiply to 8 (the previous product) and add to the second coefficient -3?



To find these two numbers, we need to list all of the factors of 8 (the previous product).



Factors of 8:

1,2,4,8

-1,-2,-4,-8



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



These factors pair up and multiply to 8.

1*8 = 8
2*4 = 8
(-1)*(-8) = 8
(-2)*(-4) = 8


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



First NumberSecond NumberSum
181+8=9
242+4=6
-1-8-1+(-8)=-9
-2-4-2+(-4)=-6




From the table, we can see that there are no pairs of numbers which add to -3. So 2x%5E2-3x%2B4 cannot be factored.



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Answer:



So 2%2Ax%5E2-3%2Ax%2B4 doesn't factor at all (over the rational numbers).



So 2%2Ax%5E2-3%2Ax%2B4 is prime.