The calculator will find all possible rational roots of the polynomial using the rational zeros theorem. After this, it will decide which possible roots are actually the roots. This is a more general case of the integer (integral) root theorem (when the leading coefficient is $$$1$$$ or $$$-1$$$). Steps are available.

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Since all coefficients are integers, we can apply the rational zeros theorem.

The trailing coefficient (the coefficient of the constant term) is $$$7$$$.

Find its factors (with the plus sign and the minus sign): $$$\pm 1$$$, $$$\pm 7$$$.

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 $$$\frac{p}{q}$$$: $$$\pm \frac{1}{1}$$$, $$$\pm \frac{1}{2}$$$, $$$\pm \frac{7}{1}$$$, $$$\pm \frac{7}{2}$$$.

Simplify and remove the duplicates (if any).

These are the possible rational roots: $$$\pm 1$$$, $$$\pm \frac{1}{2}$$$, $$$\pm 7$$$, $$$\pm \frac{7}{2}$$$.

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 $$$2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$$$ by $$$x - 1$$$.

$$$P{\left(1 \right)} = -12$$$; thus, the remainder is $$$-12$$$.

Check $$$-1$$$: divide $$$2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$$$ by $$$x - \left(-1\right) = x + 1$$$.

$$$P{\left(-1 \right)} = 0$$$; thus, the remainder is $$$0$$$.

Hence, $$$-1$$$ is a root.

Check $$$\frac{1}{2}$$$: divide $$$2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$$$ by $$$x - \frac{1}{2}$$$.

$$$P{\left(\frac{1}{2} \right)} = 0$$$; thus, the remainder is $$$0$$$.

Hence, $$$\frac{1}{2}$$$ is a root.

Check $$$- \frac{1}{2}$$$: divide $$$2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$$$ by $$$x - \left(- \frac{1}{2}\right) = x + \frac{1}{2}$$$.

$$$P{\left(- \frac{1}{2} \right)} = \frac{27}{4}$$$; thus, the remainder is $$$\frac{27}{4}$$$.

Check $$$7$$$: divide $$$2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$$$ by $$$x - 7$$$.

$$$P{\left(7 \right)} = 4368$$$; thus, the remainder is $$$4368$$$.

Check $$$-7$$$: divide $$$2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$$$ by $$$x - \left(-7\right) = x + 7$$$.

$$$P{\left(-7 \right)} = 3780$$$; thus, the remainder is $$$3780$$$.

Check $$$\frac{7}{2}$$$: divide $$$2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$$$ by $$$x - \frac{7}{2}$$$.

$$$P{\left(\frac{7}{2} \right)} = \frac{567}{4}$$$; thus, the remainder is $$$\frac{567}{4}$$$.

Check $$$- \frac{7}{2}$$$: divide $$$2 x^{4} + x^{3} - 15 x^{2} - 7 x + 7$$$ by $$$x - \left(- \frac{7}{2}\right) = x + \frac{7}{2}$$$.

$$$P{\left(- \frac{7}{2} \right)} = 105$$$; thus, the remainder is $$$105$$$.

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**Possible rational roots: $$$\pm 1$$$, $$$\pm \frac{1}{2}$$$, $$$\pm 7$$$, $$$\pm \frac{7}{2}$$$A.**