If one of the zeroes of the cubic polynomial x 3 +ax 2 +bx+c is -1, then the product of other two zeroes is:

The correct answer is C, and the product of the other two zeroes is c.

Let the zeroes be -1, p and q. For a monic cubic x^3 + ax^2 + bx + c, the product of all three zeroes equals -c, the negative of the constant term. So (-1) x p x q = -c, which gives pq = c. The minus sign from the cubic formula and the minus sign from the zero -1 cancel, so the product is positive c.

Options involving a or b are tempting because they come from the sum of zeroes (-a) and the sum of pairwise products (b). Those relate to different combinations of roots, so they do not answer this question. Likewise, -c is the product of all three zeroes, not the other two.

Core concept: Vieta's formulas for polynomials. This is a standard algebra tool for Prelims, and the same reasoning appears in polynomial-based problems in quantitative and reasoning sections.

Trap: Candidates write -c directly, forgetting to divide out the known zero -1 from the three-zero product.