College Algebra, Chapter 4, 4.4, Section 4.4, Problem 54

Determine all the real zeros of the polynomial $P(x) = 4x^5 - 18x^4 - 6x^3 + 91x^2 - 60x + 9$. Use the quadratic formula if necessary.

The leading coefficient of $P$ is $4$, and its factors are $\pm 1, \pm 2, \pm 4$. They are the divisors of constant term $9$ and its factors are $\pm 1, \pm 3, \pm 9$. Thus, the possible zeros are

$\displaystyle \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{9}{4}$

Using Synthetic Division,







We find that $1$ is not a zero but that $3$ is a zero and that $P$ factors as

$\displaystyle 4x^5 - 18x^4 - 6x^3 + 91x^2 - 60x + 9 = (x - 3) \left( 4x^4 - 6x^3 - 24x^2 + 19x - 3 \right)$

We now factor the quotient $4x^4 - 6x^3 - 24x^2 + 19x - 3$ and the possible zeros are

$\displaystyle \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}$

Using Synthetic Division,







We find that $-1$ is not a zero but that $3$ is a zero and that $P$ factors as

$\displaystyle 4x^5 - 18x^4 - 6x^3 + 91x^2 - 60x + 9 = (x - 3)(x - 3) \left( 4x^3 - 6x^2 - 6x + 1 \right)$

We now factor the quotient $4x^3 + 6x^2 - 6x + 1$ and the possible zeros are

$\displaystyle \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}$

Using Synthetic Division,







We find that $\displaystyle \frac{1}{2}$ is a zero and that $P$ factors as

$\displaystyle 4x^5 - 18x^4 - 6x^3 + 91x^2 - 60x + 9 = (x - 3)^2 \left( x - \frac{1}{2} \right) (4x^2 + 8x - 2)$

We now factor the quotient $4x^2 + 8x - 2$ using quadratic formula


$\begin{equation} \begin{aligned} x =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \\ x =& \frac{- (8) \pm \sqrt{(8)^2 - 4 (4)(-2)}}{2 (4)} \\ \\ x =& \frac{-2 \pm \sqrt{6}}{2} \end{aligned} \end{equation}$


The zeros of $P$ are $\displaystyle 3, \frac{1}{2}, \frac{-2 + \sqrt{6}}{2}$ and $\displaystyle \frac{-2 - \sqrt{6}}{2}$.