a) Ta có: F(x)+Q(x)=x5−x3+2.
Suy ra Q(x)=x5−x3+2−F(x)=x5−x3+2−(x7−12x3+x+1)
=x5−x3+2−x7+12x3−x−1=−x7+x5−12x3−x+1
Vậy Q(x) = −x7+x5−12x3−x+1
b) Ta có F(x) - R(x) = 2.
Suy ra R(x) = F(x) - 2 = x7−12x3+x+1−2=x7−12x3+x−1