\(\left(x^4+ax^3+b\right)\) \(⋮\) \(\left(x^2-1\right)\)
\(\Leftrightarrow\left(x^4+ax^3+b\right)\) \(⋮\) \(\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^4+ax^3+b\right)⋮\left(x-1\right)\\\left(x^4+ax^3+b\right)⋮\left(x+1\right)\end{matrix}\right.\)
Đặt \(f\left(x\right)=x^4+ax^3+b\)
Theo định lý Bezout , số dư của \(f\left(x\right)=x^4+ax^3+b\) cho \(\left\{{}\begin{matrix}x-1\\x+1\end{matrix}\right.\)là
\(\left\{{}\begin{matrix}f\left(-1\right)=x^4+ax^3+b=1-a+b=0\left(1\right)\\f\left(1\right)=x^4+ax^3+b=1+a+b=0\end{matrix}\right.\)
\(\Leftrightarrow1-a+b=1+a+b\)
\(\Leftrightarrow2a=0\)
\(\Leftrightarrow a=0\)
Thay \(a=0\) vào phương trình \(\left(1\right)\)
\(\Rightarrow1+b=0\)
\(\Leftrightarrow b=-1\)
Vậy \(\left\{{}\begin{matrix}a=0\\b=-1\end{matrix}\right.\)
