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Possible and actual rational roots of $$$f{\left(x \right)} = x^{3} - 6 x^{2} + 3 x + 10$$$
Your Input
Find the rational zeros of $$$x^{3} - 6 x^{2} + 3 x + 10 = 0$$$.
Solution
Since all coefficients are integers, we can apply the rational zeros theorem.
The trailing coefficient (the coefficient of the constant term) is $$$10$$$.
Find its factors (with the plus sign and the minus sign): $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 5$$$, $$$\pm 10$$$.
These are the possible values for $$$p$$$.
The leading coefficient (the coefficient of the term with the highest degree) is $$$1$$$.
Find its factors (with the plus sign and the minus sign): $$$\pm 1$$$.
These are the possible values for $$$q$$$.
Find all possible values of $$$\frac{p}{q}$$$: $$$\pm \frac{1}{1}$$$, $$$\pm \frac{2}{1}$$$, $$$\pm \frac{5}{1}$$$, $$$\pm \frac{10}{1}$$$.
Simplify and remove the duplicates (if any).
These are the possible rational roots: $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 5$$$, $$$\pm 10$$$.
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 $$$x^{3} - 6 x^{2} + 3 x + 10$$$ by $$$x - 1$$$.
$$$P{\left(1 \right)} = 8$$$; thus, the remainder is $$$8$$$.
Check $$$-1$$$: divide $$$x^{3} - 6 x^{2} + 3 x + 10$$$ by $$$x - \left(-1\right) = x + 1$$$.
$$$P{\left(-1 \right)} = 0$$$; thus, the remainder is $$$0$$$.
Hence, $$$-1$$$ is a root.
Check $$$2$$$: divide $$$x^{3} - 6 x^{2} + 3 x + 10$$$ by $$$x - 2$$$.
$$$P{\left(2 \right)} = 0$$$; thus, the remainder is $$$0$$$.
Hence, $$$2$$$ is a root.
Check $$$-2$$$: divide $$$x^{3} - 6 x^{2} + 3 x + 10$$$ by $$$x - \left(-2\right) = x + 2$$$.
$$$P{\left(-2 \right)} = -28$$$; thus, the remainder is $$$-28$$$.
Check $$$5$$$: divide $$$x^{3} - 6 x^{2} + 3 x + 10$$$ by $$$x - 5$$$.
$$$P{\left(5 \right)} = 0$$$; thus, the remainder is $$$0$$$.
Hence, $$$5$$$ is a root.
Check $$$-5$$$: divide $$$x^{3} - 6 x^{2} + 3 x + 10$$$ by $$$x - \left(-5\right) = x + 5$$$.
$$$P{\left(-5 \right)} = -280$$$; thus, the remainder is $$$-280$$$.
Check $$$10$$$: divide $$$x^{3} - 6 x^{2} + 3 x + 10$$$ by $$$x - 10$$$.
$$$P{\left(10 \right)} = 440$$$; thus, the remainder is $$$440$$$.
Check $$$-10$$$: divide $$$x^{3} - 6 x^{2} + 3 x + 10$$$ by $$$x - \left(-10\right) = x + 10$$$.
$$$P{\left(-10 \right)} = -1620$$$; thus, the remainder is $$$-1620$$$.
Answer
Possible rational roots: $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 5$$$, $$$\pm 10$$$A.
Actual rational roots: $$$-1$$$, $$$2$$$, $$$5$$$A.