## Precalculus (6th Edition) Blitzer

$x=-1, x=4, x=\frac{3+\sqrt{13}}{2}, x=\frac{3-\sqrt{13}}{2}$
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 $x=\frac{3+\sqrt{13}}{2}$.