College Algebra, Chapter 4, 4.4, Section 4.4, Problem 30

Determine all rational zeros of the polynomial $P(x) = 2x^4 - x^3 - 19x^2 + 9x + 9$, and write the polynomial in factored form.

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

Using the rational zeros theorem

possible rational zero of $P = \displaystyle \frac{\text{factor of 9}}{\text{factor of 2}}$

The possible rational zeros of $P$ are

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

Simplifying the fractions, we get

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

Using Synthetic Division







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

$2x^4 - x^3 - 19x^2 + 9x + 9 = (x - 1)(2x^3 + x^2 - 18x - 9)$

We now factor the quotient $2x^3 + x^2 - 18x - 9$. The factors of $9$ are $\pm 1, \pm 3, \pm 9$ and the factors of $2$ are $\pm 1, \pm 2$. The possible rational zeros are

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

Using Synthetic Division







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

$2x^4 - x^3 - 19x^2 + 9x + 9 = (x - 1)(x - 3)(2x^2 + 7x + 3)$

We now factor $2x^2 + 7x + 3$ using trial and error, so we get

$2x^4 - x^3 - 19x^2 + 9x + 9 = (x - 1) (x - 3) (2x + 1) (x + 3)$

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