Possible and actual rational roots of $$$f{\left(x \right)} = x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$

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 $$$1$$$.

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 $$$\frac{p}{q}$$$: $$$\pm \frac{1}{1}$$$, $$$\pm \frac{2}{1}$$$, $$$\pm \frac{3}{1}$$$, $$$\pm \frac{6}{1}$$$.

Simplify and remove the duplicates (if any).

These are the possible rational roots: $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 3$$$, $$$\pm 6$$$.

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 $$$x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$ by $$$x - 1$$$.

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

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

• Check $$$-1$$$: divide $$$x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$ by $$$x - \left(-1\right) = x + 1$$$.

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

• Check $$$2$$$: divide $$$x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$ by $$$x - 2$$$.

  $$$P{\left(2 \right)} = 10$$$; thus, the remainder is $$$10$$$.

• Check $$$-2$$$: divide $$$x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$ by $$$x - \left(-2\right) = x + 2$$$.

  $$$P{\left(-2 \right)} = -6$$$; thus, the remainder is $$$-6$$$.

• Check $$$3$$$: divide $$$x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$ by $$$x - 3$$$.

  $$$P{\left(3 \right)} = 84$$$; thus, the remainder is $$$84$$$.

• Check $$$-3$$$: divide $$$x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$ by $$$x - \left(-3\right) = x + 3$$$.

  $$$P{\left(-3 \right)} = 0$$$; thus, the remainder is $$$0$$$.

  Hence, $$$-3$$$ is a root.

• Check $$$6$$$: divide $$$x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$ by $$$x - 6$$$.

  $$$P{\left(6 \right)} = 1530$$$; thus, the remainder is $$$1530$$$.

• Check $$$-6$$$: divide $$$x^{4} + 2 x^{3} - 5 x^{2} - 4 x + 6$$$ by $$$x - \left(-6\right) = x + 6$$$.

  $$$P{\left(-6 \right)} = 714$$$; thus, the remainder is $$$714$$$.

Possible rational roots: $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 3$$$, $$$\pm 6$$$A.

Actual rational roots: $$$1$$$, $$$-3$$$A.