Question

(Mackenzie-SP) A equação kx³+x²+1=0 , com k E Z, admite raiz inteira.Então, k pode ser:
a) -1 b)-3 c)3 d)1 e)2

Answer 1

sea 

$\displaystyle a\in \mathbb Z\\ \\ (x-a)(kx^2+rx-1/a) =kx^3 + (r - ak)x^2 - \frac{a^2r + 1}{a}x + 1\\ \\ r-ak=1 \Longrightarrow \boxed{k=\frac{r-1}{a}}\\ \\ \frac{a^2r + 1}{a}=0\Longrightarrow \boxed{r=-\frac{1}{a^2}}\\ \\ k=-\frac{a^2+1}{a^3}\\ \\ \text{adem\'as: si }a\ \textgreater \ 1 \text{ entonces }a^3\ \textgreater \ a^2+1\ \textgreater \ 0 \iff 0\ \textless \ \frac{a^2+1}{a^3}\ \textless \ 1\\ \\ \text{Si } a\ \textless \ -1\Longrightarrow 0\ \textless \ -\frac{a^2+1}{a^3}\ \textless \ 1\\ \\ \text{Por lo tanto }a\in[-1,1] \text{ y como }a\in \mathbb Z \Longrightarrow a\in\{-1,1\}$

entonces

$\large\boxed{k\in\{2,-2\}}$

Respuesta (e)