# How do you find all the zeros of g(x) = 9x^3- 7x^2 + 10x - 4?

Aug 11, 2016

Use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{27} \left(7 + \sqrt[3]{- 1882 + 81 \sqrt{2185}} + \sqrt[3]{- 1882 - 81 \sqrt{2185}}\right)$

and related Complex zeros.

#### Explanation:

$g \left(x\right) = 9 {x}^{3} - 7 {x}^{2} + 10 x - 4$

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Descriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

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

In our example, $a = 9$, $b = - 7$, $c = 10$ and $d = - 4$, so we find:

$\Delta = 4900 - 36000 - 5488 - 34992 + 45360 = - 26220$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 2187 g \left(x\right) = 19683 {x}^{3} - 15309 {x}^{2} + 21870 x - 8748$

$= {\left(27 x - 7\right)}^{3} + 663 \left(27 x - 7\right) - 3764$

$= {t}^{3} + 663 t - 3764$

where $t = \left(27 x - 7\right)$

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Cardano's method

We want to solve:

${t}^{3} + 663 t - 3764 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v + 221\right) \left(u + v\right) - 3764 = 0$

Add the constraint $v = - \frac{221}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} - \frac{10793861}{u} ^ 3 - 3764 = 0$

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

${\left({u}^{3}\right)}^{2} - 3764 \left({u}^{3}\right) - 10793861 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{3764 \pm \sqrt{{\left(- 3764\right)}^{2} - 4 \left(1\right) \left(- 10793861\right)}}{2 \cdot 1}$

$= \frac{- 3764 \pm \sqrt{14167696 + 43175444}}{2}$

$= \frac{- 3764 \pm \sqrt{57343140}}{2}$

$= \frac{- 3764 \pm 162 \left(2185\right)}{2}$

$= - 1882 \pm 81 \left(2185\right)$

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

${t}_{1} = \sqrt[3]{- 1882 + 81 \sqrt{2185}} + \sqrt[3]{- 1882 - 81 \sqrt{2185}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{- 1882 + 81 \sqrt{2185}} + {\omega}^{2} \sqrt[3]{- 1882 - 81 \sqrt{2185}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{- 1882 + 81 \sqrt{2185}} + \omega \sqrt[3]{- 1882 - 81 \sqrt{2185}}$

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

Now $x = \frac{1}{27} \left(7 + t\right)$. So the zeros of our original cubic are:

${x}_{1} = \frac{1}{27} \left(7 + \sqrt[3]{- 1882 + 81 \sqrt{2185}} + \sqrt[3]{- 1882 - 81 \sqrt{2185}}\right)$

${x}_{2} = \frac{1}{27} \left(7 + \omega \sqrt[3]{- 1882 + 81 \sqrt{2185}} + {\omega}^{2} \sqrt[3]{- 1882 - 81 \sqrt{2185}}\right)$

${x}_{3} = \frac{1}{27} \left(7 + {\omega}^{2} \sqrt[3]{- 1882 + 81 \sqrt{2185}} + \omega \sqrt[3]{- 1882 - 81 \sqrt{2185}}\right)$