If (x-6)(x+2) = x² + bx + c for all values of x, what is the value of b - c?

Expand: $(x - 6)(x + 2) = x^2 + 2x - 6x - 12 = x^2 - 4x - 12$. So $b = -4$ and $c = -12$. The question asks for $b - c = -4 - (-12) = -4 + 12 = 8$, choice B. Choice A ($-16$) is $b + c$. Choice C ($-4$) is $b$ alone. Choice D ($-8$) has the wrong sign, which comes from computing $c - b$ or from subtracting without handling the double negative. Careful sign work is the whole challenge here, so write out $b - c$ with the values in parentheses.