Let p1(x)=x3−2020x2+b1x+c1=(x−α)(x−β)(x−γ) and p2(x)=x3−2021x2+b2x+c2=(x−α)(x−β)(x−δ) ∵ p1(x).q1(x)+p2(x)q2(x)=x2−3x+2 Comparing the coefficient of x3 we get q1(x)=−q2(x)=q(x)( say ) So (x−α)(x−β)[q(x)(δ−γ)]=(x−1)(x−2) Hence α=1,β=2,γ=2017 and δ=2018 p1(x)=(x−1)(x−2)(x−2017)⇒p1(3)=−4028 p2(x)=(x−1)(x−2)(x−2018) So p1(3)+p2(1)+4028=0