Browse · MATH Print → jmc algebra intermediate Problem Let a, b, c be the roots of x3−7x+2=0. Find abc+ab+ac+bc+a+b+c+1. Solution — click to reveal By Vieta's formulas, a+b+c=0, ab+ac+bc=−7, and abc=−2, so abc+ab+ac+bc+a+b+c+1=(−2)+(−7)+0+1=−8. Final answer -8 ← Previous problem Next problem →