# What are the zeros of f(x) = x^3-140x^2+7984x-107584?

Nov 2, 2016

Use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{3} \left(140 + 8 \sqrt[3]{1628 + 12 \sqrt{20589}} + 8 \sqrt[3]{1628 - 12 \sqrt{20589}}\right)$

and two related Complex zeros.

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} - 140 {x}^{2} + 7984 x - 107584$

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Discriminant

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, $a = 1$, $b = - 140$, $c = 7984$ and $d = - 107584$, so we find:

$\Delta = 1249387417600 - 2035736559616 - 1180841984000 - 312506560512 + 2164555653120 = - 115142033408$

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.

$0 = 27 f \left(x\right) = 27 {x}^{3} - 3780 {x}^{2} + 215568 x - 2904768$

$= {\left(3 x - 140\right)}^{3} + 13056 \left(3 x - 140\right) + 1667072$

$= {t}^{3} + 13056 t + 1667072$

where $t = \left(3 x - 140\right)$

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Cardano's method

We want to solve:

${t}^{3} + 13056 t + 1667072 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v + 4352\right) \left(u + v\right) + 1667072 = 0$

Add the constraint $v = - \frac{4352}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} - \frac{82426462208}{u} ^ 3 + 1667072 = 0$

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

${\left({u}^{3}\right)}^{2} + 1667072 \left({u}^{3}\right) - 82426462208 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{- 1667072 \pm \sqrt{{\left(1667072\right)}^{2} - 4 \left(1\right) \left(- 82426462208\right)}}{2 \cdot 1}$

$= \frac{1667072 \pm \sqrt{2779129053184 + 329705848832}}{2}$

$= \frac{1667072 \pm \sqrt{3108834902016}}{2}$

$= \frac{1667072 \pm 12288 \sqrt{20589}}{2}$

$= 833536 \pm 6144 \sqrt{20589}$

$= {8}^{3} \left(1628 \pm 12 \sqrt{20589}\right)$

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:

${t}_{1} = 8 \sqrt[3]{1628 + 12 \sqrt{20589}} + 8 \sqrt[3]{1628 - 12 \sqrt{20589}}$

and related Complex roots:

${t}_{2} = 8 \omega \sqrt[3]{1628 + 12 \sqrt{20589}} + 8 {\omega}^{2} \sqrt[3]{1628 - 12 \sqrt{20589}}$

${t}_{3} = 8 {\omega}^{2} \sqrt[3]{1628 + 12 \sqrt{20589}} + 8 \omega \sqrt[3]{1628 - 12 \sqrt{20589}}$

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

Now $x = \frac{1}{3} \left(140 + t\right)$. So the roots of our original cubic are:

${x}_{1} = \frac{1}{3} \left(140 + 8 \sqrt[3]{1628 + 12 \sqrt{20589}} + 8 \sqrt[3]{1628 - 12 \sqrt{20589}}\right)$

${x}_{2} = \frac{1}{3} \left(140 + 8 \omega \sqrt[3]{1628 + 12 \sqrt{20589}} + 8 {\omega}^{2} \sqrt[3]{1628 - 12 \sqrt{20589}}\right)$

${x}_{3} = \frac{1}{3} \left(140 + 8 {\omega}^{2} \sqrt[3]{1628 + 12 \sqrt{20589}} + 8 \omega \sqrt[3]{1628 - 12 \sqrt{20589}}\right)$

Nov 2, 2016

$x \approx 18.9 , x \approx 60 + 45 i , \mathmr{and} x \approx 60 - 45 i$