SOLUTION: 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!

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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 NumberSecond NumberSum
1-321+(-32)=-31
2-162+(-16)=-14
4-84+(-8)=-4
-132-1+32=31
-216-2+16=14
-48-4+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): You can put this solution on YOUR website!
For to be factorable using integers, it has to be factorable using rational numbers. The condition is, the discriminant has to be a perfect square rational number. In your question, a = 8, b=-3, and c = -4.
==>
which is not a perfect square rational number. Thus it cannot be factored using rational numbers, and hence not using integers.
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