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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