If α and β are the roots of $x^2-4x+7=0$, then (α − 1)(β − 1) equals:

Expand first: (α − 1)(β − 1) = αβ − (α + β) + 1 = 7 − 4 + 1 = 4. (2 drops the sign on the final 1, giving 7 − 4 − 1; 12 adds everything as 7 + 4 + 1; 10 mixes the two slips.) There is a neat shortcut: for $p(x)=x^2-4x+7$ the product (α − 1)(β − 1) is exactly $p(1)=1-4+7=4$, and it works even though the roots 2 ± i√3 are not real.