SOLUTION: i seem to be having trouble with this problem. ive done everything i could 2h^2-3h-18 thank you for your time

Question 179494: i seem to be having trouble with this problem. ive done everything i could
2h^2-3h-18
thank you for your time
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Do you want to factor? Please post full instructions.

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 :
1,2,3,4,6,9,12,18,36
-1,-2,-3,-4,-6,-9,-12,-18,-36

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

These factors pair up and multiply to .
1*(-36)
2*(-18)
3*(-12)
4*(-9)
6*(-6)
(-1)*(36)
(-2)*(18)
(-3)*(12)
(-4)*(9)
(-6)*(6)

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	-36	1+(-36)=-35
2	-18	2+(-18)=-16
3	-12	3+(-12)=-9
4	-9	4+(-9)=-5
6	-6	6+(-6)=0
-1	36	-1+36=35
-2	18	-2+18=16
-3	12	-3+12=9
-4	9	-4+9=5
-6	6	-6+6=0

From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.

So I would make sure that you entered the correct problem.
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