hhhmm :> Làm cho bạn 2 pp còn lại tự túc ak <33
Đặt \(f\left(x\right)=x^3+ax+b\)
C1 :
Có \(x^2-x-2=\left(x+1\right)\left(x-2\right)\)
\(f\left(x\right)⋮\left(x^2-x-2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(-1\right)=0\\f\left(2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1-a+b=0\\8+2a+b=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b-a=1\\2a+b=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b-a-2a-b=-3a=9\\b-a=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-2\end{matrix}\right.\)
C2 : Giả sử
\(f\left(x\right)=x^3+ax+b=\left(x^2-x-2\right)\left(x+c\right)\)
\(=x^3-x^2-2x+cx^2-cx-2c\)
\(=x^3+x^2\left(c-1\right)+x\left(-2-c\right)-2c\)
Áp dụng đồng nhất thức ta có
\(\left\{{}\begin{matrix}c=1\\-2-c=a\\-2c=b\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-2\end{matrix}\right.\)
