# What are all the zeroes of f(x)=5x^3+8x^2-4x+3?

Nov 22, 2015

The Real root is:

${x}_{1} = - \frac{8}{15} + \sqrt[3]{\frac{- 4489 + 45 \sqrt{6185}}{6750}} + \sqrt[3]{\frac{- 4489 - 45 \sqrt{6185}}{6750}}$

$\approx - 2.11299545715260$

Complex roots as shown below.

#### Explanation:

$f \left(x\right) = 5 {x}^{3} + 8 {x}^{2} - 4 x + 3$

This is of the form $a {x}^{3} + b {x}^{2} + c x + d$ with $a = 5$, $b = 8$, $c = - 4$ and $d = 3$

Its discriminant $\Delta$ is given by the formula:

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

$= 1024 + 1280 - 6144 - 6075 - 8640 = - 18555$

Since $\Delta < 0$ this cubic has one Real zero and two Complex conjugate ones.

So we can use a Tschirnhaus transformation to eliminate the ${x}^{2}$ term, then use Cardano's method.

Let $t = x + \frac{8}{15}$

Then:

$5 {t}^{3} = 5 {\left(x + \frac{8}{15}\right)}^{3} = 5 {x}^{3} + 8 {x}^{2} + \frac{64}{15} x + \frac{512}{675}$

$- \frac{124}{15} t = - \frac{124}{15} \left(x + \frac{8}{15}\right) = - \frac{124}{15} x - \frac{992}{225}$

Hence:

$5 {t}^{3} - \frac{124}{15} t + \frac{4489}{675} = 5 {x}^{3} + 8 {x}^{2} - 4 x + 3$

Multiplying through by $675$, we want to solve:

$3375 {t}^{3} - 5580 t + 4489 = 0$

Let $t = u + v$

$3375 {u}^{3} + 3375 {v}^{3} + \left(10125 u v - 5580\right) \left(u + v\right) + 4489 = 0$

To eliminate the $\left(u + v\right)$ term add the constraint:

$v = \frac{5580}{10125 u} = \frac{124}{225 u}$

Then we have:

$0 = 3375 {u}^{3} + 3375 {\left(\frac{124}{225 u}\right)}^{3} + 4489$

$= 3375 {u}^{3} + 3375 \left(\frac{1906624}{11390625 {u}^{3}}\right) + 4489$

$= 3375 {u}^{3} + \frac{1906624}{3375 {u}^{3}} + 4489$

Multiply through by $3375 {u}^{3}$ to get:

$11390625 {\left({u}^{3}\right)}^{2} + 15150375 \left({u}^{3}\right) + 1906624 = 0$

${u}^{3} = \frac{- 15150375 \pm \sqrt{{15150375}^{2} - \left(4 \cdot 11390625 \cdot 1906624\right)}}{2 \cdot 11390625}$

$= \frac{- 15150375 \pm \sqrt{142663306640625}}{22781250}$

$= \frac{- 15150375 \pm 151875 \sqrt{6185}}{22781250}$

$= \frac{- 4489 \pm 45 \sqrt{6185}}{6750}$

Since the derivation is symmetric in $u$ and $v$, we can deduce:

$t = \sqrt[3]{\frac{- 4489 + 45 \sqrt{6185}}{6750}} + \sqrt[3]{\frac{- 4489 - 45 \sqrt{6185}}{6750}}$

${x}_{1} = - \frac{8}{15} + \sqrt[3]{\frac{- 4489 + 45 \sqrt{6185}}{6750}} + \sqrt[3]{\frac{- 4489 - 45 \sqrt{6185}}{6750}}$

The Complex roots are:

${x}_{2} = - \frac{8}{15} + \omega \sqrt[3]{\frac{- 4489 + 45 \sqrt{6185}}{6750}} + {\omega}^{2} \sqrt[3]{\frac{- 4489 - 45 \sqrt{6185}}{6750}}$

${x}_{3} = - \frac{8}{15} + {\omega}^{2} \sqrt[3]{\frac{- 4489 + 45 \sqrt{6185}}{6750}} + \omega \sqrt[3]{\frac{- 4489 - 45 \sqrt{6185}}{6750}}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$