36x^2+8x+K,find the value of k such that each trinomial can be factored over the integers

Answer 1

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.