Question

${x}^{3} - 5 {x}^{2} - x + 5 = 0$
resolva a equação ​

Answer 1

$\boxed{x^3-5x^2-x+5=0}$

$x=1\ \to\ (1)^3-5(1)^2-1+5=0\ \to\ \boxed{x=1}\ \mathrm{\acute{e}\ raiz}$

$\text{Dispositivo de Briot-Ruffini:}$

$\begin{array}{ccc}1\end{array}\right|\left|\begin{array}{cccc}1&-5&-1&5\\&1&-4&-5\\1&-4&-5&0\end{array}\right|$

$(x-1)(x^2-4x-5)=0\ \to\ x^2-4x-5=0$

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(-5)}}{2(1)}=$

$\dfrac{4\pm\sqrt{36}}{2}=\dfrac{4\pm6}{2}=2\pm3\ \to\ \boxed{x=5}\ \text{e}\ \boxed{x=-1}$

$\boxed{x=\left\{-1,1,5\right\}}$