# How do you find all the zeros of f(x) = 4x^3 – 5x^2 + 9x – 6?

##### 1 Answer
Mar 8, 2016

Multiply by $432$, substitute $t = 12 x - 5$, then use Cardano's method,

#### Explanation:

$f \left(x\right) = 4 {x}^{3} - 5 {x}^{2} + 9 x - 6$

First multiply through by ${3}^{3} \cdot {4}^{2} = 432$ to cut down on the number of fractions we need to work with.

$0 = 432 f \left(x\right)$

$= 1728 {x}^{3} - 2160 {x}^{2} + 3888 x - 2592$

$= {\left(12 x - 5\right)}^{3} + 249 \left(12 x - 5\right) - 1222$

Substitute $t = 12 x - 5$

${t}^{3} + 249 t - 1222 = 0$

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

${u}^{3} + {v}^{3} + \left(3 u v + 249\right) \left(u + v\right) - 1222 = 0$

Add the constraint $v = - \frac{249}{3 u}$ to eliminate the term in $\left(u + v\right)$

${u}^{3} - {249}^{3} / \left(27 {u}^{3}\right) - 1222 = 0$

Multiply through by $27 {u}^{3}$ and rearrange to get:

$27 {\left({u}^{3}\right)}^{2} - 32994 \left({u}^{3}\right) - 15438249 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{32994 \pm \sqrt{2755934928}}{54}$

$= \frac{32994 \pm 972 \sqrt{2917}}{54}$

$= 611 \pm 18 \sqrt{2917}$

Due to the symmetry of the derivation in $u$ and $v$, we can take one of these roots as the value of ${u}^{3}$ and the other as ${v}^{3}$ to get the Real root:

${t}_{1} = \sqrt[3]{611 + 18 \sqrt{2917}} + \sqrt[3]{611 - 18 \sqrt{2917}}$

and Complex roots:

${t}_{2} = \omega \sqrt[3]{611 + 18 \sqrt{2917}} + {\omega}^{2} \sqrt[3]{611 - 18 \sqrt{2917}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{611 + 18 \sqrt{2917}} + \omega \sqrt[3]{611 - 18 \sqrt{2917}}$

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

Then you can derive the roots of the original equation:

${x}_{1} = \frac{{t}_{1} + 5}{12} = \frac{1}{12} \left(5 + \sqrt[3]{611 + 18 \sqrt{2917}} + \sqrt[3]{611 - 18 \sqrt{2917}}\right)$

${x}_{2} = \frac{1}{12} \left(5 + \omega \sqrt[3]{611 + 18 \sqrt{2917}} + {\omega}^{2} \sqrt[3]{611 - 18 \sqrt{2917}}\right)$

${x}_{3} = \frac{1}{12} \left(5 + {\omega}^{2} \sqrt[3]{611 + 18 \sqrt{2917}} + \omega \sqrt[3]{611 - 18 \sqrt{2917}}\right)$