# What are all the zeroes of f(x) = 2x^3 - 5x^2 + 3x - 1?

Dec 19, 2015

#### Answer:

The Real root of $f \left(x\right) = 0$ is:

${x}_{1} = \frac{1}{6} \left(5 + \sqrt[3]{44 + 3 \sqrt{177}} + \sqrt[3]{44 - 3 \sqrt{177}}\right)$

and Complex roots as below...

#### Explanation:

Given:

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

First substitute $t = x - \frac{5}{6}$

Then:

$2 {t}^{3} = 2 {\left(x - \frac{5}{6}\right)}^{3} = 2 \left({x}^{3} - \frac{5}{2} {x}^{2} + \frac{25}{12} x - \frac{125}{216}\right)$

$= 2 {x}^{3} - 5 {x}^{2} + \frac{25}{6} x - \frac{125}{108}$

$\textcolor{w h i t e}{}$

$- \frac{7}{6} t = - \frac{7}{6} x + \frac{35}{36} = - \frac{7}{6} x + \frac{105}{108}$

$\textcolor{w h i t e}{}$

$2 {t}^{3} - \frac{7}{6} t = 2 {x}^{3} - 5 {x}^{2} + 3 x - \frac{5}{27}$

So:

$2 {t}^{3} - \frac{7}{6} t - \frac{22}{27} = 2 {x}^{3} - 5 {x}^{2} + 3 x - 1 = f \left(x\right)$

Multiply through by $54$ to get integer coefficients:

$108 {t}^{3} - 63 t - 44 = 54 f \left(x\right)$

So we want to solve:

$108 {t}^{3} - 63 t - 44 = 0$

Use Cardano's method, letting $t = u + v$

$0 = 108 {\left(u + v\right)}^{3} - 63 \left(u + v\right) - 44$

$= 108 {u}^{3} + 108 {v}^{3} + \left(324 u v - 63\right) \left(u + v\right) - 44$

$= 108 {u}^{3} + 108 {v}^{3} + 9 \left(36 u v - 7\right) \left(u + v\right) - 44$

Next make the coefficient of the $\left(u + v\right)$ term zero by adding the constraint: $36 u v - 7 = 0$, that is $v = \frac{7}{36 u}$

$= 108 {u}^{3} + 108 {\left(\frac{7}{36 u}\right)}^{3} - 44$

$= 108 {u}^{3} + \frac{343}{432 {u}^{3}} - 44$

Multiply through by $432 {u}^{3}$ to get a quadratic in ${u}^{3}$:

$46656 {\left({u}^{3}\right)}^{2} - 19008 \left({u}^{3}\right) + 343 = 0$

Then using the quadratic formula:

${u}^{3} = \frac{19008 \pm \sqrt{{19008}^{2} - 4 \cdot 46656 \cdot 343}}{2 \cdot 46656}$

$= \frac{19008 \pm \sqrt{361304064 - 64012032}}{93312}$

$= \frac{19008 \pm \sqrt{297292032}}{93312}$

$= \frac{19008 \pm 1296 \sqrt{177}}{93312}$

$= \frac{44 \pm 3 \sqrt{177}}{216}$

Since the derivation was symmetric in $u$ and $v$, one of these roots is suitable for ${u}^{3}$ and the other for ${v}^{3}$.

Hence the Real root is:

$t = \sqrt[3]{\frac{44 + 3 \sqrt{177}}{216}} + \sqrt[3]{\frac{44 - 3 \sqrt{177}}{216}}$

$= \frac{1}{6} \left(\sqrt[3]{44 + 3 \sqrt{177}} + \sqrt[3]{44 - 3 \sqrt{177}}\right)$

and hence:

${x}_{1} = \frac{5}{6} + t = \frac{1}{6} \left(5 + \sqrt[3]{44 + 3 \sqrt{177}} + \sqrt[3]{44 - 3 \sqrt{177}}\right)$

The Complex roots are:

${x}_{2} = \frac{1}{6} \left(5 + \omega \sqrt[3]{44 + 3 \sqrt{177}} + {\omega}^{2} \sqrt[3]{44 - 3 \sqrt{177}}\right)$

${x}_{3} = \frac{1}{6} \left(5 + {\omega}^{2} \sqrt[3]{44 + 3 \sqrt{177}} + \omega \sqrt[3]{44 - 3 \sqrt{177}}\right)$

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