A function is given by f(x) = (x-5)(x+2)(x-1). What is the sum of the zeros of f?

The zeros of a factored polynomial are the values that make each factor equal zero: $x-5=0$ gives $5$, $x+2=0$ gives $-2$, and $x-1=0$ gives $1$. Their sum is $5+(-2)+1=4$, choice A. Vieta's formulas agree: for a cubic $x^3+bx^2+\cdots$ the roots sum to $-b$, and expanding gives $x^3-4x^2-\cdots$, so the sum is $4$. Choice B ($-4$) is the coefficient of $x^2$, not the opposite of it. Choice C ($8$) adds the absolute values of the roots. Choice D ($-6$) is not derived from the zeros.