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

Aug 16, 2016

Use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{27} \left(1 + \sqrt[3]{\frac{6235 + 27 \sqrt{46221}}{2}} + \sqrt[3]{\frac{6235 - 27 \sqrt{46221}}{2}}\right)$

and related Complex zeros.

#### Explanation:

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

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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, $a = - 9$, $b = 1$, $c = 4$ and $d = - 3$, so we find:

$\Delta = 16 + 2304 + 12 - 19683 + 1944 = - 15407$

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.

$0 = - 2187 f \left(x\right) = 19683 {x}^{3} - 2187 {x}^{2} - 8748 x + 6561$

$= {\left(27 x - 1\right)}^{3} - 327 \left(27 x - 1\right) + 6235$

$= {t}^{3} - 327 t + 6235$

where $t = \left(27 x - 1\right)$

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Cardano's method

We want to solve:

${t}^{3} - 327 t + 6235 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 109\right) \left(u + v\right) + 6235 = 0$

Add the constraint $v = \frac{109}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{1295029}{u} ^ 3 + 6235 = 0$

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

${\left({u}^{3}\right)}^{2} + 6235 \left({u}^{3}\right) + 1295029 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{- 6235 \pm \sqrt{{\left(6235\right)}^{2} - 4 \left(1\right) \left(1295029\right)}}{2 \cdot 1}$

$= \frac{6235 \pm \sqrt{38875225 - 5180116}}{2}$

$= \frac{6235 \pm \sqrt{33695109}}{2}$

$= \frac{6235 \pm 27 \sqrt{46221}}{2}$

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:

${t}_{1} = \sqrt[3]{\frac{6235 + 27 \sqrt{46221}}{2}} + \sqrt[3]{\frac{6235 - 27 \sqrt{46221}}{2}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{\frac{6235 + 27 \sqrt{46221}}{2}} + {\omega}^{2} \sqrt[3]{\frac{6235 - 27 \sqrt{46221}}{2}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{\frac{6235 + 27 \sqrt{46221}}{2}} + \omega \sqrt[3]{\frac{6235 - 27 \sqrt{46221}}{2}}$

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

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

${x}_{1} = \frac{1}{27} \left(1 + \sqrt[3]{\frac{6235 + 27 \sqrt{46221}}{2}} + \sqrt[3]{\frac{6235 - 27 \sqrt{46221}}{2}}\right)$

${x}_{2} = \frac{1}{27} \left(1 + \omega \sqrt[3]{\frac{6235 + 27 \sqrt{46221}}{2}} + {\omega}^{2} \sqrt[3]{\frac{6235 - 27 \sqrt{46221}}{2}}\right)$

${x}_{3} = \frac{1}{27} \left(1 + {\omega}^{2} \sqrt[3]{\frac{6235 + 27 \sqrt{46221}}{2}} + \omega \sqrt[3]{\frac{6235 - 27 \sqrt{46221}}{2}}\right)$