Question 397336: I need to factor this trinomial completely, I thinks it is "not factorable" using integers but I want to be sure.
8t^2 - 3t - 4
Thanks in advance!
Found 2 solutions by jim_thompson5910, robertb:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Looking at  we can see that the first term is  and the last term is  where the coefficients are 8 and -4 respectively.
Now multiply the first coefficient 8 and the last coefficient -4 to get -32. Now what two numbers multiply to -32 and add to the middle coefficient -3? Let's list all of the factors of -32:
Factors of -32:
1,2,4,8,16,32
-1,-2,-4,-8,-16,-32 ...List the negative factors as well. This will allow us to find all possible combinations
These factors pair up and multiply to -32
(1)*(-32)
(2)*(-16)
(4)*(-8)
(-1)*(32)
(-2)*(16)
(-4)*(8)
note: remember, the product of a negative and a positive number is a negative number
Now which of these pairs add to -3? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -3
| First Number | Second Number | Sum | | 1 | -32 | 1+(-32)=-31 | | 2 | -16 | 2+(-16)=-14 | | 4 | -8 | 4+(-8)=-4 | | -1 | 32 | -1+32=31 | | -2 | 16 | -2+16=14 | | -4 | 8 | -4+8=4 |
None of these pairs of factors add to -3. So the expression cannot be factored
So you are correct. Good job.
Answer by robertb(5830)  (Show Source):
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