College Algebra, Chapter 4, 4.4, Section 4.4, Problem 50

Determine all the real zeros of the polynomial $P(x) = x^5 - 4x^4 - x^3 + 10x^2 + 2x - 4$. Use the quadratic formula if necessary.

The leading coefficient of $P$ is $1$, so all the rational zeros are integers. They are the divisors of constant term $-4$. Thus, the possible candidates are

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

Using Synthetic Division,







We find that $1$ is not a zero but that $2$ is a zero and that $P$ factors as

$\displaystyle x^5 - 4x^4 - x^3 + 10x^2 + 2x - 4 = (x - 2) \left( x^4 - 2x^3 - 5x^2 + 2 \right)$

We now factor the quotient $x^4 - 2x^3 - 5x^2 + 2$ and its possible zeros are

$ \pm 1, \pm 2$

Using Synthetic Division,







We find that $-1$ is a zero and that $P$ factors as

$\displaystyle x^5 - 4x^4 - x^3 + 10x^2 + 2x - 4 = (x - 2)(x + 1) \left( x^3 - 3x^2 - 2x + 2 \right)$

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

$\pm 1, \pm 2$

Using Synthetic Division,







We find that $-1$ is a zero and that $P$ factors as

$\displaystyle x^5 - 4x^4 - x^3 + 10x^2 + 2x - 4 = (x - 2)(x + 1)(x + 1) (x^2 - 4x + 2)$

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


$
\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 $-1, 2, 2 + \sqrt{2}$ and $2 - \sqrt{2}$.