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Possible and actual rational roots of $$$f{\left(x \right)} = x^{3} + 2 x^{2} - 19 x - 20$$$

The calculator will find all possible and actual rational roots of the polynomial $$$f{\left(x \right)} = x^{3} + 2 x^{2} - 19 x - 20$$$, with steps shown.

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Your Input

Find the rational zeros of $$$x^{3} + 2 x^{2} - 19 x - 20 = 0$$$.

Solution

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

The trailing coefficient (the coefficient of the constant term) is $$$-20$$$.

Find its factors (with the plus sign and the minus sign): $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 4$$$, $$$\pm 5$$$, $$$\pm 10$$$, $$$\pm 20$$$.

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 $$$\frac{p}{q}$$$: $$$\pm \frac{1}{1}$$$, $$$\pm \frac{2}{1}$$$, $$$\pm \frac{4}{1}$$$, $$$\pm \frac{5}{1}$$$, $$$\pm \frac{10}{1}$$$, $$$\pm \frac{20}{1}$$$.

Simplify and remove the duplicates (if any).

These are the possible rational roots: $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 4$$$, $$$\pm 5$$$, $$$\pm 10$$$, $$$\pm 20$$$.

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 $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - 1$$$.

    $$$P{\left(1 \right)} = -36$$$; thus, the remainder is $$$-36$$$.

  • Check $$$-1$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ 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 $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - 2$$$.

    $$$P{\left(2 \right)} = -42$$$; thus, the remainder is $$$-42$$$.

  • Check $$$-2$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - \left(-2\right) = x + 2$$$.

    $$$P{\left(-2 \right)} = 18$$$; thus, the remainder is $$$18$$$.

  • Check $$$4$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - 4$$$.

    $$$P{\left(4 \right)} = 0$$$; thus, the remainder is $$$0$$$.

    Hence, $$$4$$$ is a root.

  • Check $$$-4$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - \left(-4\right) = x + 4$$$.

    $$$P{\left(-4 \right)} = 24$$$; thus, the remainder is $$$24$$$.

  • Check $$$5$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - 5$$$.

    $$$P{\left(5 \right)} = 60$$$; thus, the remainder is $$$60$$$.

  • Check $$$-5$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - \left(-5\right) = x + 5$$$.

    $$$P{\left(-5 \right)} = 0$$$; thus, the remainder is $$$0$$$.

    Hence, $$$-5$$$ is a root.

  • Check $$$10$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - 10$$$.

    $$$P{\left(10 \right)} = 990$$$; thus, the remainder is $$$990$$$.

  • Check $$$-10$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - \left(-10\right) = x + 10$$$.

    $$$P{\left(-10 \right)} = -630$$$; thus, the remainder is $$$-630$$$.

  • Check $$$20$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - 20$$$.

    $$$P{\left(20 \right)} = 8400$$$; thus, the remainder is $$$8400$$$.

  • Check $$$-20$$$: divide $$$x^{3} + 2 x^{2} - 19 x - 20$$$ by $$$x - \left(-20\right) = x + 20$$$.

    $$$P{\left(-20 \right)} = -6840$$$; thus, the remainder is $$$-6840$$$.

Answer

Possible rational roots: $$$\pm 1$$$, $$$\pm 2$$$, $$$\pm 4$$$, $$$\pm 5$$$, $$$\pm 10$$$, $$$\pm 20$$$A.

Actual rational roots: $$$-1$$$, $$$4$$$, $$$-5$$$A.