Short answer: A <<<<< answer.
I'm taking this to mean that all terms have a coefficient that is an integer.
y = 2x^2 + kx + 5
One way to to do this is to look at the discriminate in the quadratic equation.
b = k
c = 5
a = 2
y1 = sqrt(b^2 - 4ac)
y1 = sqrt(k^2 - 4(2)(5))
y1 = sqrt(k^2 - 40)
Why do that? For one thing it lets out a couple of answers. C and D for 1 will will give complex numbers, which means that the real roots cannot be found.
sqrt(1^2 - 40) = sqrt(-39) which produces nothing useful. Both C and D will not work.
B is not much more promising. the 2 gives trouble.
square root (2^2 - 40) = sqrt(-36) so that's out.
We have only one more to try. Start with 7
sqrt(7^2 - 40) = sqrt(49 - 40) = 9. That looks good.
x = [-7 +/- sqrt(49 - 4*2*5)] / 2*2
x = [-7 +/- sqrt (9)]/4
x = [-7 +/- 3] / 4
x = -10/4 = -5/2
Does it factor?
2x^2 - 7 x + 5
y = (2x - 5)(x - 1)
There are two results for 7
x = [-7 +/- sqrt(49 - 40)]/4
x = [-7 + 3] / 4
x = -1
When k = -1, does this make 2x^2 - x + 5 factorable.
No but we've already found 1 value that works.
What about when k = 11?
2x^2 + 11x + 5 = y
(2x + 1)(x + 5) = y and yes 11 creates a factorable number.
Answer A <<<<