f(x)=x4−2x3+3x2−ax+b Given that, f(1)=5 and f(−1)=19 , then f(1)=1−2+3−a+b=2−a+b=5 ∴b−a=3 ...(i) and f(−1)=1+2+3+a+b=b+a+6=19 ∴b+a=13 ...(ii) adding Eqs. (i) and (ii), we obtain 2b=16⇒b=8 anda=13−b⇒a=5 ∴f(x)=x4−2x3+3x2−5x+8 Thus, f(2)=(2)4−2(2)3+3(2)2−5(2)+8 =16−16+12−10+8=10 ∴ If f(x) is divided by x−2 , then remainder is 10 .