We will work with options here (since the question clearly asks - which of the "following" could be odd)
Let x = 2k, where k is an integer (since x is even, it must be a multiple of 2)
Option A: \((x^2 + x)(x + 2)/4\) = \((4k^2 + 2k)(2k + 2)/4\) = \(2k(2k + 1) * 2(k + 1)/4\) = \(k(2k + 1) * (k + 1)\)
Here, the first term (2k + 1) is odd; however, k(k + 1) is always even
Thus, Option A is even
Option B: \(x^3 - 3x^2 + 2x\) = x(x - 1)(x - 2)
If x is even, the value is even
Thus, Option B is even
Option C: \(x^2 - 4x + 6\) = (x - 2)(x - 2) + 2
If x is even, the value is (even) * (even) + 2, which is even
Thus, Option C is even
Option D: \((x^3 + x^2)/4\) = \((8k^3 + 4k^2)/4\) = \((2k^3 + k^2)\)
If k is odd, the value is odd
Thus, Option D COULD be odd
Option E: \((x^2 + 2x)(x - 2)/4\) = \((4k^2 + 4k)(2k - 2)/4\) = \(4k(2k + 1) * 2(k - 1)/4\) = \(2k(2k + 1) * (k - 1)\)
This is clearly even
Thus, Option E is even
Answer D