# How do you find all the zeros of x^3+2x^2-3x-3?

Jun 8, 2016

Use a trigonometric solution to find:

${x}_{k} = \frac{1}{3} \left(- 2 + 2 \sqrt{13} \cos \left(\frac{1}{3} \left(\arccos \left(\frac{11 \sqrt{13}}{338}\right) + 2 k \pi\right)\right)\right)$

$k = 0 , 1 , 2$

#### Explanation:

$f \left(x\right) = {x}^{3} + 2 {x}^{2} - 3 x - 3$

Using the rational root theorem we find that the only possible rational zeros are:

$\pm 1$, $\pm 3$

None of these work, so none of the zeros are rational.

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 1$, $b = 2$, $c = - 3$ and $d = - 3$, so we find:

$\Delta = 36 + 108 + 96 - 243 + 324 = 321$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

As a result, pure algebraic methods like Cardano's method will yield results containing irreducible cube roots of Complex numbers. So choose to use a trigonometric solution instead.

First, use a Tschirnhaus transformation to eliminate the square term...

$27 f \left(x\right) = 27 {x}^{3} + 54 {x}^{2} - 81 x - 81 = {\left(3 x + 2\right)}^{3} - 39 \left(3 x + 2\right) - 11$

Let $t = \left(3 x + 2\right)$

We want to solve:

${t}^{3} - 39 t - 11 = 0$

Next, use a substitution of the form:

$t = 2 \sqrt{13} \cos \theta$

The multiplier $2 \sqrt{13}$ is chosen so that our cubic contains a multiple of $4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$ as follows:

$0 = {\left(2 \sqrt{13} \cos \theta\right)}^{3} - 39 \left(2 \sqrt{13} \cos \theta\right) - 11$

$= 26 \sqrt{13} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 11$

$= 26 \sqrt{13} \cos 3 \theta - 11$

So:

$\cos 3 \theta = \frac{11}{26 \sqrt{13}} = \frac{11 \sqrt{13}}{338}$

$\theta = \pm \frac{1}{3} \left(\arccos \left(\frac{11 \sqrt{13}}{338}\right) + 2 k \pi\right)$

$\cos \theta = \cos \left(\frac{1}{3} \left(\arccos \left(\frac{11 \sqrt{13}}{338}\right) + 2 k \pi\right)\right)$

$t = 2 \sqrt{13} \cos \left(\frac{1}{3} \left(\arccos \left(\frac{11 \sqrt{13}}{338}\right) + 2 k \pi\right)\right)$

${x}_{k} = \frac{1}{3} \left(- 2 + 2 \sqrt{13} \cos \left(\frac{1}{3} \left(\arccos \left(\frac{11 \sqrt{13}}{338}\right) + 2 k \pi\right)\right)\right)$

which has distinct values for $k = 0 , 1 , 2$

${x}_{0} \approx 1.4605048700187635$

${x}_{1} \approx - 2.699628148275318$

${x}_{2} \approx - 0.7608767217434455$