If the zeroes of the polynomial \(f(x)=x^3+px^2+qx-6\) are \(1\), \(2\), and \(3\), what is the value of \(p+q\)?

Correct answer: 5

Using the given zeroes, the polynomial can be written as \((x-1)(x-2)(x-3)\). On expansion, this becomes \(x^3-6x^2+11x-6\). Comparing coefficients gives \(p=-6\) and \(q=11\), so \(p+q=-6+11=5\). Exam tip: For a monic cubic polynomial, compare the coefficients of \(x^2\) and \(x\) directly after forming the product of the corresponding linear factors.