Question

Tìm a để :

(a³- a²- 4a +1 ) : ( a +1 ) là số nguyên

Answer 1

\(\dfrac{\left(a^3-a^2-4a+1\right)}{\left(a+1\right)}=\dfrac{\left(a+1\right)\left(a^2-2a-2\right)+3}{a+1}\)

\(\dfrac{\left(a^3-a^2-4a+1\right)}{\left(a+1\right)}=\left(a^2-2a-2\right)+\dfrac{3}{a+1}\)

để \(\dfrac{\left(a^3-a^2-4a+1\right)}{\left(a+1\right)}\) là số nguyên \(\Leftrightarrow3⋮\left(a+1\right)\)

\(\Leftrightarrow\left(a+1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow\left[{}\begin{matrix}a+1=1\\a+1=-1\\a+1=3\\a+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=-2\\a=2\\a=-4\end{matrix}\right.\)