When will ((n-4)x^2+(n+1)x+6) become a constant polynomial?

Correct answer: Never

The direct answer is D, never. For an expression to be constant in x, every term containing x must disappear. Here the coefficient of x^2 is n-4, so we need n-4=0, giving n=4. The coefficient of x is n+1, so we also need n+1=0, giving n=-1. One value of n cannot be both 4 and -1. The constant term 6 is already independent of x. Option A, n=4, removes the x^2 term, but the x coefficient becomes 5, so the expression still has an x term. Option B, n=-1, removes the x term, but the x^2 coefficient becomes -5, so an x^2 term remains. Option C says n=4 and n=-1 together; that is impossible for one parameter, so it cannot occur. Option D is correct because no single value satisfies both necessary equations. The reasoning is not that constants are impossible; rather, this particular expression cannot become constant for one common value of n. Exam cue: set every variable-term coefficient equal to zero, then check whether the resulting conditions are compatible.