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Possible and actual rational roots of $$$f{\left(x \right)} = 2 x^{3} - 6 x^{2} - x + 6$$$
Your Input
Find the rational zeros of $$$2 x^{3} - 6 x^{2} - x + 6 = 0$$$.
Solution
Since all coefficients are integers, we can apply the rational zeros theorem.
The trailing coefficient (the coefficient of the constant term) is $$$6$$$.
Find its factors (with the plus sign and the minus sign): $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 3$$$, $$$\pm 6$$$.
These are the possible values for $$$p$$$.
The leading coefficient (the coefficient of the term with the highest degree) is $$$2$$$.
Find its factors (with the plus sign and the minus sign): $$$\pm 1$$$, $$$\pm 2$$$.
These are the possible values for $$$q$$$.
Find all possible values of $$$\frac{p}{q}$$$: $$$\pm \frac{1}{1}$$$, $$$\pm \frac{1}{2}$$$, $$$\pm \frac{2}{1}$$$, $$$\pm \frac{2}{2}$$$, $$$\pm \frac{3}{1}$$$, $$$\pm \frac{3}{2}$$$, $$$\pm \frac{6}{1}$$$, $$$\pm \frac{6}{2}$$$.
Simplify and remove the duplicates (if any).
These are the possible rational roots: $$$\pm 1$$$, $$$\pm \frac{1}{2}$$$, $$$\pm 2$$$, $$$\pm 3$$$, $$$\pm \frac{3}{2}$$$, $$$\pm 6$$$.
Next, check the possible roots: if $$$a$$$ is a root of the polynomial $$$P{\left(x \right)}$$$, the remainder from the division of $$$P{\left(x \right)}$$$ by $$$x - a$$$ should equal $$$0$$$ (according to the remainder theorem, this means that $$$P{\left(a \right)} = 0$$$).
Check $$$1$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - 1$$$.
$$$P{\left(1 \right)} = 1$$$; thus, the remainder is $$$1$$$.
Check $$$-1$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - \left(-1\right) = x + 1$$$.
$$$P{\left(-1 \right)} = -1$$$; thus, the remainder is $$$-1$$$.
Check $$$\frac{1}{2}$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - \frac{1}{2}$$$.
$$$P{\left(\frac{1}{2} \right)} = \frac{17}{4}$$$; thus, the remainder is $$$\frac{17}{4}$$$.
Check $$$- \frac{1}{2}$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - \left(- \frac{1}{2}\right) = x + \frac{1}{2}$$$.
$$$P{\left(- \frac{1}{2} \right)} = \frac{19}{4}$$$; thus, the remainder is $$$\frac{19}{4}$$$.
Check $$$2$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - 2$$$.
$$$P{\left(2 \right)} = -4$$$; thus, the remainder is $$$-4$$$.
Check $$$-2$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - \left(-2\right) = x + 2$$$.
$$$P{\left(-2 \right)} = -32$$$; thus, the remainder is $$$-32$$$.
Check $$$3$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - 3$$$.
$$$P{\left(3 \right)} = 3$$$; thus, the remainder is $$$3$$$.
Check $$$-3$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - \left(-3\right) = x + 3$$$.
$$$P{\left(-3 \right)} = -99$$$; thus, the remainder is $$$-99$$$.
Check $$$\frac{3}{2}$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - \frac{3}{2}$$$.
$$$P{\left(\frac{3}{2} \right)} = - \frac{9}{4}$$$; thus, the remainder is $$$- \frac{9}{4}$$$.
Check $$$- \frac{3}{2}$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - \left(- \frac{3}{2}\right) = x + \frac{3}{2}$$$.
$$$P{\left(- \frac{3}{2} \right)} = - \frac{51}{4}$$$; thus, the remainder is $$$- \frac{51}{4}$$$.
Check $$$6$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - 6$$$.
$$$P{\left(6 \right)} = 216$$$; thus, the remainder is $$$216$$$.
Check $$$-6$$$: divide $$$2 x^{3} - 6 x^{2} - x + 6$$$ by $$$x - \left(-6\right) = x + 6$$$.
$$$P{\left(-6 \right)} = -636$$$; thus, the remainder is $$$-636$$$.
Answer
Possible rational roots: $$$\pm 1$$$, $$$\pm \frac{1}{2}$$$, $$$\pm 2$$$, $$$\pm 3$$$, $$$\pm \frac{3}{2}$$$, $$$\pm 6$$$A.
Actual rational roots: no rational roots.