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

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) (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 , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

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

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

Factors of :

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 .

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 :

First Number	Second Number	Sum
1	8	1+8=9
2	4	2+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 . So cannot be factored.

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

Answer:

So doesn't factor at all (over the rational numbers).

So is prime.

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