36x^2+8x+K,find the value of k such that each trinomial can be factored over the integers
to have integer factors, we must have factors of the following forms, with k and n integers:
(36x+36k)(x-n)
(18x+18k)(2x-2n)
(12x+12k)(3x-3n)
(9x+9k)(4x-4n)
(6x+6k)(6x-6n)
Each of these multiplies out to
36x^2 + 36(k-n)x - 36kn
no way can 36(k-n)=8 with k and n integers.
I don't see any solutions.
Alternatively, using the quadratic formula,
x = (-8±√64-144k)/72 = -1/9 ± 1/18 √(4-9k)
since the roots are symmetric about -1/9, there is no way both roots can be integers.
To determine the value of k such that the trinomial 36x^2 + 8x + K can be factored over the integers, we need to find two integers factors of 36K that add up to 8.
First, let's factorize 36. The factors of 36 are:
1 and 36
2 and 18
3 and 12
4 and 9
6 and 6
Now, we need to find two factors whose sum equals 8. By examining the pairs, we see that 4 and 9 can be added to give 13, which is not what we're looking for. Therefore, we can eliminate that possibility.
Next, we try the pair 2 and 18. When we add them together, we get 20, which is still not 8.
Proceeding to the next pair, 3 and 12. Adding them gives us 15, which is also not 8.
Lastly, we have the pair 6 and 6. Adding them together gives us 12, which is still not 8.
Since none of the factor pairs add up to 8, it means there is no integer value of K that would allow the trinomial 36x^2 + 8x + K to be factored over the integers.