College Algebra, Chapter 4, 4.4, Section 4.4, Problem 40

Determine all rational zeros of the polynomial $P(x) = 6x^4 - 7x^3 - 12x^2 + 3x + 2$, and write the polynomial in factored form.

The leading coefficient of $P$ is $6$ and its factors are $\pm 1, \pm 2, \pm 3, \pm 6$. They are the divisors of the constant term $2$ and its factors are $\pm 1, \pm 2$. The possible rational zeros are $\displaystyle \pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{2}{3}$

Using Synthetic Division







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

$6x^4 - 7x^3 - 12x^2 + 3x + 2 = (x - 2)(6x^3 + 5x^2 - 2x - 1)$

We now factor the quotient $6x^3 + 5x^2 - 2x -1$. The factors of 6 are $\pm 1,\pm 2, \pm 3, \pm 6$. The factors of 1 are $\pm 1$. The possible rational zeros are
$\displaystyle \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$
Using synthetic division



We find that $-1$ is a zero and $P$ factors as
$6x^4 - 7x^3 -12x^2 + 3x + 2 = (x-2)(x+1)(6x^2 - x - 1)$

We now factor $6x^2 - x - 1$ using trial and error. We get,

$6x^4 - 7x^3 - 12x^2 + 3x + 2 = (x - 2)(x + 1)(6x + 1)(x - 1)$

The zeros of $P$ are $2, 1, -1$ and $\displaystyle \frac{-1}{6}$.