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
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a) A can be rewritten in the form \(\frac{x(x+1)(x+2)}{4}\) i.e. 3 consecutive integers -> Since x is even then the (x+2) will be divisible by 4 (whenever it is possible; won't be when x is -2 or 0 but the expression won't be odd in such cases) and the expression will be even.

b) \(x^3−3x^2+2x\) = \(x*(x^2-3x+2)\) = \(x*(x-1)*(x-2)\) -> Again multiplication of 3 consecutive integers -> Always even

c) \(x^2-4x+6\) -> Since x is even this term will always be even as even*even = even and even + even = even

d) \(\frac{x^3+x^2}{4}\) = \(\frac{x^2*(x+1)}{4}\) -> \(x^2\) will always be even and will always be divisible by 4 since x is a multiple of 2. x+1 will always be odd since x is even. Hence this could be ODD ---> Answer

e) \(\frac{(x-2)(x^2+2x)}{4}\) = \(\frac{(x-2)x(x+2)}{4}\) -> again since x is even x+2 will be divisible by 4 as in case a)
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