College Algebra, Chapter 4, 4.4, Section 4.4, Problem 48

Determine all the real zeros of the polynomial $P(x) = x^4 + 2x^3 - 2x^2 - 3x + 2$. 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 $2$. Thus, the possible candidates are

$\pm 1, \pm 2$

Using Synthetic Division,







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

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

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

$\pm 1, \pm 2$

Using Synthetic Division,







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

$x^4 + 2x^3 - 2x^2 - 3x + 2 = (x - 1)(x + 2)(x^2 + x - 1)$

We now factor the quotient $x^2 + x - 1$ using the quadratic formula


$\begin{equation} \begin{aligned} x =& \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \\ x =& \frac{-1 \pm \sqrt{(1)^2 - 4 (1)(-1)}}{2(1)} \\ \\ x =& \frac{-1 \pm \sqrt{5}}{2} \end{aligned} \end{equation}$


The zeros of $P$ are $\displaystyle 1, -2, \frac{-1 + \sqrt{5}}{2}$ and $\displaystyle \frac{-1 - \sqrt{5}}{2}$