The cubic $2x^3-3x^2-11x+6$ has 3 as one of its zeroes. If the other two zeroes are α and β, what is α + β?

All three zeroes add to $-\frac{b}{a}=\frac{3}{2}$. Removing the known zero: $\alpha +\beta =\frac{3}{2}-3=-\frac{3}{2}$. (The full set of zeroes is 3, $\frac{1}{2}$ and −2, and $\frac{1}{2}+(-2)=-\frac{3}{2}.)\frac{3}{2}$ forgets to subtract the known zero; $\frac{9}{2}$ adds 3 instead of subtracting; $-\frac{11}{2}$ reads off $\frac{c}{a}$, the pairwise sum, by mistake.