If the zeroes of the polynomial \(p(x)=x^3+px^2+qx-6\) 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 \(x^3-6x^2+11x-6\). Therefore, \(p=-6\) and \(q=11\), so \(p+q=-6+11=5\). The value \(-5\) results from subtracting instead of adding the coefficients. In an exam, form the factors from the zeroes, expand them, and compare coefficients of like powers.