# How do you find all the zeros of x^3 + 2x^2 + 5x +1?

Mar 12, 2016

Use Cardano's method...

#### Explanation:

$f \left(x\right) = {x}^{3} + 2 {x}^{2} + 5 x + 1$

To cut down on the number of fractions we need to work with, multiply by ${3}^{3} = 27$ first...

$0 = 27 f \left(x\right)$

$= 27 {x}^{3} + 54 {x}^{2} + 135 x + 27$

$= {\left(3 x + 2\right)}^{3} + 33 \left(3 x + 2\right) - 47$

Substitute $t = 3 x + 2$ ...

$= {t}^{3} + 33 t - 47$

Using Cardano's method, substitute $t = u + v$ ...

$= {u}^{3} + {v}^{3} + 3 \left(u v + 11\right) \left(u + v\right) - 47$

Add the constraint $v = - \frac{11}{u}$ to eliminate the $\left(u + v\right)$ term ...

$= {u}^{3} - \left({11}^{3} / {u}^{3}\right) - 47$

$= {u}^{3} - \frac{1331}{u} ^ 3 - 47$

Multiply through by ${u}^{3}$ to get this quadratic in ${u}^{3}$ ...

${\left({u}^{3}\right)}^{2} - 47 \left({u}^{3}\right) - 1331 = 0$

Solve using the quadratic formula to get:

${u}^{3} = \frac{47 \pm \sqrt{{47}^{2} + 4 \cdot 1331}}{2}$

$= \frac{47 \pm \sqrt{7533}}{2}$

$= \frac{47 \pm 9 \sqrt{93}}{2}$

Since the derivation was symmetric in $u$ and $v$, we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to derive the Real root:

${t}_{1} = \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$

and the Complex roots:

${t}_{2} = \omega \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + {\omega}^{2} \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + \omega \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$

Then $x = \frac{t - 2}{3}$ hence zeros of the original function:

x_1 = 1/3(-2+root(3)((47+9sqrt(93))/2) + root(3)((47-9sqrt(93))/2) )

${x}_{2} = \frac{1}{3} \left(- 2 + \omega \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + {\omega}^{2} \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left(- 2 + {\omega}^{2} \sqrt[3]{\frac{47 + 9 \sqrt{93}}{2}} + \omega \sqrt[3]{\frac{47 - 9 \sqrt{93}}{2}}\right)$