# How do you find all the real and complex roots of 3x^3+2x^2+x-4=0?

May 24, 2016

Use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{9} \left(- 2 + \sqrt{505 + 135 \sqrt{14}} + \sqrt{505 - 135 \sqrt{14}}\right)$

and related Complex zeros.

#### Explanation:

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

This cubic has no rational zeros. It turns out to have one Real zero and a Complex conjugate pair of non-Real zeros.

To simplify it while avoiding some fraction arithmetic, first multiply it by ${3}^{2} \cdot {3}^{3} = 243$ to find:

$243 f \left(x\right) = 729 {x}^{3} + 486 {x}^{2} + 243 x - 972$

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

$= {\left(9 x + 2\right)}^{3} + 15 \left(9 x + 2\right) - 1010$

Let $t = 9 x + 2$

We want to solve:

${t}^{3} + 15 t - 1010 = 0$

This substitution to reduce a polynomial to one of simpler form is called a Tschirnhaus transformation.

To solve the simplified cubic, use Cardano's method, letting $t = u + v$

${u}^{3} + {v}^{3} + 3 \left(u v + 5\right) \left(u + v\right) - 1010 = 0$

Substitute $v = - \frac{5}{u}$ to eliminate the term in $\left(u + v\right)$ and find:

${u}^{3} - \frac{125}{u} ^ 3 - 1010 = 0$

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

${\left({u}^{3}\right)}^{2} - 1010 \left({u}^{3}\right) - 125 = 0$

Using the quadratic formula, we find:

${u}^{3} = \frac{1010 \pm \sqrt{{1010}^{2} - \left(4 \cdot 1 \cdot - 125\right)}}{2 \cdot 1}$

$= \frac{1010 \pm \sqrt{1020100 + 500}}{2}$

$= \frac{1010 \pm \sqrt{1020600}}{2}$

$= 505 \pm 135 \sqrt{14}$

Since this is Real and the derivation was symmetric in $u$ and $v$, we can use one of these values as ${u}^{3}$ and the other as ${v}^{3}$ to find the Real root of ${t}^{3} + 15 t - 1010 = 0$ is:

${t}_{1} = \sqrt{505 + 135 \sqrt{14}} + \sqrt{505 - 135 \sqrt{14}}$

and Complex roots:

${t}_{2} = \omega \sqrt{505 + 135 \sqrt{14}} + {\omega}^{2} \sqrt{505 - 135 \sqrt{14}}$

${t}_{3} = {\omega}^{2} \sqrt{505 + 135 \sqrt{14}} + \omega \sqrt{505 - 135 \sqrt{14}}$

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

Then $x = \frac{1}{9} \left(t - 2\right)$, so the zeros of the original cubic are:

${x}_{1} = \frac{1}{9} \left(- 2 + \sqrt{505 + 135 \sqrt{14}} + \sqrt{505 - 135 \sqrt{14}}\right)$

${x}_{2} = \frac{1}{9} \left(- 2 + \omega \sqrt{505 + 135 \sqrt{14}} + {\omega}^{2} \sqrt{505 - 135 \sqrt{14}}\right)$

${x}_{3} = \frac{1}{9} \left(- 2 + {\omega}^{2} \sqrt{505 + 135 \sqrt{14}} + \omega \sqrt{505 - 135 \sqrt{14}}\right)$