document.write( "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... \r
\n" ); document.write( "\n" ); document.write( "4x^2-6x+8\r
\n" ); document.write( "\n" ); document.write( "If I could get the factors I know how to do the box but I can't get it factored. \n" ); document.write( "

Algebra.Com's Answer #388773 by richwmiller(17219) $\"\"$ $\"About$

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4x^2-6x+8
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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 $\"-3\"$ :

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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 $\"-3\"$ . 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.

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