College Algebra, Chapter 4, 4.4, Section 4.4, Problem 60

The polynomial $P(x) = -x^4 + 10x^2 + 8x - 8$.

a.) Find all the real zeros of $P$

The leading coefficient of $P$ is $-1$, so all rational zeros are integers. They are divisors of constant term $-8$. Thus, the possible zeros are

$\displaystyle \pm 1, \pm 2, \pm 4, \pm 8$

Using Synthetic Division







We find that $1, 2, 4$ and $-1$ are not zeros but that $-2$ is zero and that $P$ factors as

$\displaystyle -x^4 + 10x^2 + 8x - 8 = (x + 2)\left( -x^3 + 2x^2 + 6x - 4 \right)$

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

$\pm 1, \pm 2, \pm 4$

Using Synthetic Division







We find that $-2$ is a zero and that $P$ factors as

$\displaystyle -x^4 + 10x^2 + 8x - 8 = (x + 2) (x + 2) (-x^2 + 4x - 2)$

We now factor the quotient $-x^2 + 4x - 2$ using Quadratic Formula, we get


$
\begin{equation}
\begin{aligned}

x =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\
\\
x =& \frac{-4 \pm \sqrt{(4)^2 - 4(-1)(-2)}}{2 (-1)}
\\
\\
x =& 2 \pm \sqrt{2}


\end{aligned}
\end{equation}
$


The zeros of $P$ are $-2, 2 + \sqrt{2}$ and $2 - \sqrt{2}$.

b.) Sketch the graph of $P$