Answer
$\frac{1}{2}$, $3$, and $\frac{-2\pm \sqrt{6}}{2}$
Work Step by Step
$P(x)=0$
$4x^5-18x^4-6x^3+91x^2-60x+9=0$ (Factorize)
$(2x-1)(x-3)(x-3)(2x^2+4x-1)=0$
Find all real zeros:
(i) $2x-1=0\to x=\frac{1}{2}$
(ii) $(x-3)=0\to x=3$
(iii) $2x^2+4x-1=0\to x=\frac{-4\pm \sqrt{4^2-4\cdot 2\cdot (-1)}}{2\cdot 2}=\frac{-4\pm 2\sqrt{6}}{4}=\frac{-2\pm \sqrt{6}}{2}$
Therefore, the real zeros of $P$ are $\frac{1}{2}$, $3$, and $\frac{-2\pm \sqrt{6}}{2}$.
