College Algebra, Chapter 4, 4.4, Section 4.4, Problem 44

Determine all rational zeros of the polynomial $P(x) = 2x^6 - 3x^5 - 13x^4 + 29x^3 -27x^2 + 32x - 12$, and write the polynomial in factored form.

The leading coefficient of $P$ is $2$ and its factors ara $\pm 1, \pm 2$. They are divisors of the constant term $-12$ and its factors are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$. Thus, the possible rational zeros are $\displaystyle \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}$.

Using Synthetic Division







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

$\displaystyle 2x^6 - 3x^5 - 13x^4 + 29x^3 - 27x^2 + 32x - 12 = (x - 2) \left( 2x^5 + x^4 - 11x^3 + 7x^2 - 13x + 6 \right)$

We now factor the quotient $2x^5 + x^4 -11 x^3 + 7x^2 - 13x + 6$. Its possible zeros are the divisors of $6$, namely

$\displaystyle \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$

Using Synthetic Division







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

$\displaystyle 2x^6 - 3x^5 - 13x^4 + 29x^3 - 27x^2 + 32x - 12 = (x - 2)(x - 2) \left(2x^4 + 5x^3 - x^2 + 5x - 3 \right)$

We now factor the quotient $2x^4 + 5x^3 - x^2 + 5x - 3$. Its possible zeros are the divisors of $-3$, namely

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

Using Synthetic Division







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

$\displaystyle 2x^6 - 3x^5 - 13x^4 + 29x^3 - 27x^2 + 32x - 12 = (x - 2)(x - 2) \left( x - \frac{1}{2} \right) \left( 2x^3 + 6x^2 + 2x + 6 \right)$

We now factor the quotient $2x^3 + 6x^2 + 2x + 69$. Its possible zeros are the divisors of $6$, namely

$\displaystyle \pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$







We find that $-2$ is not a zero but that $-3$ is a zero and that $P$ factors as


$
\begin{equation}
\begin{aligned}

2x^6 - 3x^5 - 13x^4 + 29x^3 - 27x^2 + 32x - 12 =& (x - 2)(x -2) \left( x - \frac{1}{2} \right) (x + 3) (2x^2 + 2)
\\
\\
2x^6 - 3x^5 - 13x^4 + 29x^3 - 27x^2 + 32x - 12 =& 2 (x - 2)(x - 2) \left( x - \frac{1}{2} \right) (x + 3) (x^2 + 1)

\end{aligned}
\end{equation}
$