Answer
(a) The solutions are $x = 1, -2$
(b) The solutions are $x = -3, 6$
(c) The solutions are $x = -1$ and two more complex roots.
In all three cases, using Cardano's formula seems more complicated than using methods we learned in this section.
Work Step by Step
(a) $x^3-3x+2 = 0$
$p = -3$
$q = 2$
We can use the formula to find a solution for the equation:
$x = \sqrt[3] {\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3] {\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$
$x = \sqrt[3] {\frac{-2}{2}+\sqrt{\frac{2^2}{4}+\frac{(-3)^3}{27}}}+\sqrt[3] {\frac{-2}{2}-\sqrt{\frac{2^2}{4}+\frac{(-3)^3}{27}}}$
$x = \sqrt[3] {-1+\sqrt{0}}+\sqrt[3] {-1-\sqrt{0}}$
$x = (-1)+(-1)$
$x = -2$
To solve this question using methods in this section, we can list the possible rational zeros:
$\frac{factors~of~2}{factors~of~1} = \pm 1, \pm 2$
We can use synthetic division:
$1 \vert ~~1~~~0 ~~-3~~~~2$
$~~~~~~~~~~~1~~~~1~~~-2$
$~~~~1~~~1~~~-2~~~0$
The remainder is zero so $x=1$ is a solution.
We can factor:
$x^3-3x+2 = 0$
$(x-1)(x^2+x-2) = 0$
$(x-1)(x+2)(x-1) = 0$
The solutions are $x = 1, -2$
(b) $x^3-27x-54 = 0$
$p = -27$
$q = -54$
We can use the formula to find a solution for the equation:
$x = \sqrt[3] {\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3] {\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$
$x = \sqrt[3] {\frac{54}{2}+\sqrt{\frac{(-54)^2}{4}+\frac{(-27)^3}{27}}}+\sqrt[3] {\frac{54}{2}-\sqrt{\frac{(-54)^2}{4}+\frac{(-27)^3}{27}}}$
$x = \sqrt[3] {27+\sqrt{0}}+\sqrt[3] {27-\sqrt{0}}$
$x = (3)+(3)$
$x = 6$
To solve this question using methods in this section, we can list the possible rational zeros:
$\frac{factors~of~54}{factors~of~1} = \pm 1, \pm 2, \pm 3, \pm6, \pm 9, \pm 18, \pm 27, \pm 54$
We can use synthetic division:
$-3 \vert ~~1~~~0 ~~-27~~~~-54$
$~~~~~~~~~~~-3~~~~9~~~54$
$~~~~1~~~-3~~~-18~~~0$
The remainder is zero so $x=-3$ is a solution.
We can factor:
$x^3-27x-54 = 0$
$(x+3)(x^2-3x-18) = 0$
$(x+3)(x+3)(x-6) = 0$
The solutions are $x = -3, 6$
(c) $x^3+3x+4 = 0$
$p = 3$
$q = 4$
We can use the formula to find a solution for the equation:
$x = \sqrt[3] {\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3] {\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$
$x = \sqrt[3] {\frac{-4}{2}+\sqrt{\frac{4^2}{4}+\frac{3^3}{27}}}+\sqrt[3] {\frac{-4}{2}-\sqrt{\frac{4^2}{4}+\frac{3^3}{27}}}$
$x = \sqrt[3] {-2+\sqrt{5}}+\sqrt[3] {-2-\sqrt{5}}$
$x = -1$
To solve this question using methods in this section, we can list the possible rational zeros:
$\frac{factors~of~4}{factors~of~1} = \pm 1, \pm 2, \pm 4$
We can use synthetic division:
$-1 \vert ~~1~~~0 ~~3~~~~4$
$~~~~~~~~~~~-1~~~~1~~~-4$
$~~~~1~~~-1~~~4~~~0$
The remainder is zero so $x=-1$ is a solution.
We can factor:
$x^3+3x+4 = 0$
$(x+1)(x^2-x+4) = 0$
Note that $x^2-x+4$ has no real roots.
The solutions are $x = -1$ and two more complex roots.
In all three cases, using Cardano's formula seems more complicated than using methods we learned in this section.
