College Algebra, Chapter 4, 4.4, Section 4.4, Problem 36

Determine all rational zeros of the polynomial $P(x) = 8x^3 + 10x^2 - x - 3$, and write the polynomial in factored form.

The leading coefficient of $P$ is $8$ and the factors of $8$ are $\pm 1, \pm 2, \pm 4, \pm 8$. They are the divisors of the constant term $-3$ and its factors are $\pm 1, \pm 3$. The possible rational zeros are $\displaystyle \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{1}{8}, \pm \frac{3}{8}$

Using Synthetic Division







We find that $1$ and $3$ are not zeros but that $\displaystyle \frac{1}{2}$ is a zero and that $P$ factors as

$\displaystyle 8x^3 + 10x^2 - x - 3 = \left(x - \frac{1}{2} \right) \left( 8x^2 + 14x + 6 \right)$

We now factor the quotient $8x^2 + 14x + 6$ using trial and error. We get,


$\begin{equation} \begin{aligned} 8x^3 + 10x^2 - x - 3 =& \left(x - \frac{1}{2} \right) (4x + 3)(2x + 2) \\ \\ 8x^3 + 10x^2 - x - 3 =& 2 \left( x - \frac{1}{2} \right) (4x + 3)(x + 1) \end{aligned} \end{equation}$


The zeros of $P$ are $\displaystyle \frac{1}{2}, \frac{-3}{4}$ and $-1$.