Question

Xác định hệ số a, b để f(x) chia hết cho g(x), với:

\(f\left(x\right)=x^4+4\)

\(g\left(x\right)=x^2+ax+b\)

Answer 1

Ta có : \(f\left(x\right)=x^4+4=\left(x^4+4x^2+4\right)-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)

Để \(f\left(x\right)\) chia hết cho \(g\left(x\right)\) thì :

\(\left[{}\begin{matrix}x^2+2x+2⋮x^2+ax+b\\x^2-2x+2⋮x^2+ax+b\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=2\end{matrix}\right.\\\left\{{}\begin{matrix}a=-2\\b=2\end{matrix}\right.\end{matrix}\right.\)