Answer
$x=-1, x=4, x=\frac{3+\sqrt{13}}{2}, x=\frac{3-\sqrt{13}}{2}$
Work Step by Step
1) Factor the equation:
$x^{4}-6x^{3}+4x^{2}+15x+4=0$
$\Rightarrow(x+1)(x-4)(x^{2}-3x-1)=0$.
2) Solve for the roots of the equation.
Set each factor equal to zero and solve for $x$:
$(x+1)=0$
$\Rightarrow \fbox{$x=-1$}$.
$(x-4)=0$
$\Rightarrow \fbox{$x=4$}$.
$(x^{2}-3x-1)=0$
Apply the quadratic formula where $a=1$, $b=-3$, and $c=-1$:
$x=\frac{-(-3)\pm\sqrt{(-3)^{2}-4(1)(-1)}}{2(1)}\Rightarrow x=\fbox{$\frac{3\pm\sqrt{13}}{2}$}$.
Therefore, the solutions to the equation $x^{4}-6x^{3}+4x^{2}+15x+4=0$ are $x=-1, x=4, x=\frac{3+\sqrt{13}}{2},$ and $ x=\frac{3-\sqrt{13}}{2}$.
