(x3+ax+b)\(⋮\)(x-2) thì:
f(2) = 0
f(2) =23+a.2+b = 0
= 8 +2a +b = 0
\(\Rightarrow\)2a+b = -8 (✳)
(x3+ax+b)\(⋮\)(x+1) thì:
f(-1) = 0
f(-1)= (-1)3+a.(-1)+b = 0
= -1 -a +b = 0
= -a+b = 1 (✳✳)
\(\left\{{}\begin{matrix}2a+b=-8\\-a+b=1\end{matrix}\right.\)
3a = -9
a = -3
(-3)+b =1
\(\Rightarrow\) 3+b = 1
b = 1-3
b = -2 .