# How do you find the zeros of x^5 - 5x^4 - x^3 + x^2 + 4 = 0?

Aug 7, 2016

Zeros:

${x}_{1} = 1$

${x}_{2} = - 1$

${x}_{3} = \frac{1}{3} \left(5 + \sqrt[3]{179 + 12 \sqrt{114}} + \sqrt[3]{179 - 12 \sqrt{114}}\right)$

and two related Complex zeros.

#### Explanation:

${x}^{5} - 5 {x}^{4} - {x}^{3} + {x}^{2} + 4 = 0$

First note that the sum of the coefficients is zero.

That is:

$1 - 5 - 1 + 1 + 4 = 0$

So ${x}_{1} = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{5} - 5 {x}^{4} - {x}^{3} + {x}^{2} + 4 = \left(x - 1\right) \left({x}^{4} - 4 {x}^{3} - 5 {x}^{2} - 4 x - 4\right)$

The sum of the coefficients of the remaining quartic with signs reversed on terms of odd degree is zero.

That is:

$1 + 4 - 5 + 4 - 4 = 0$

So ${x}_{2} = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{4} - 4 {x}^{3} - 5 {x}^{2} - 4 x - 4 = \left(x + 1\right) \left({x}^{3} - 5 {x}^{2} - 4\right)$

By the rational root theorem, any rational zeros of the remaining cubic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 4$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4$

None of these work, so the cubic only has irrational zeros.

Let: $f \left(x\right) = {x}^{3} - 5 {x}^{2} - 4$

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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, $a = 1$, $b = - 5$, $c = 0$ and $d = - 4$, so we find:

$\Delta = 0 + 0 - 2000 - 432 + 0 = - 2432$

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.

$0 = 27 f \left(x\right) = 27 {x}^{3} - 135 {x}^{2} - 108$

$= {\left(3 x - 5\right)}^{3} - 75 \left(3 x - 5\right) - 358$

$= {t}^{3} - 75 t - 358$

where $t = \left(3 x - 5\right)$

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Cardano's method

We want to solve:

${t}^{3} - 75 t - 358 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 25\right) \left(u + v\right) - 358 = 0$

Add the constraint $v = \frac{25}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{15625}{u} ^ 3 - 358 = 0$

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

${\left({u}^{3}\right)}^{2} - 358 \left({u}^{3}\right) + 15625 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{358 \pm \sqrt{{\left(- 358\right)}^{2} - 4 \left(1\right) \left(15625\right)}}{2 \cdot 1}$

$= \frac{- 358 \pm \sqrt{128164 - 62500}}{2}$

$= \frac{- 358 \pm \sqrt{65664}}{2}$

$= \frac{- 358 \pm 24 \sqrt{114}}{2}$

$= - 119 \pm 12 \sqrt{114}$

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:

${t}_{3} = \sqrt[3]{- 119 + 12 \sqrt{114}} + \sqrt[3]{- 119 - 12 \sqrt{114}}$

and related Complex roots:

${t}_{4} = \omega \sqrt[3]{- 119 + 12 \sqrt{114}} + {\omega}^{2} \sqrt[3]{- 119 - 12 \sqrt{114}}$

${t}_{5} = {\omega}^{2} \sqrt[3]{- 119 + 12 \sqrt{114}} + \omega \sqrt[3]{- 119 - 12 \sqrt{114}}$

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

Now $x = \frac{1}{3} \left(5 + t\right)$. So the zeros of $f \left(x\right)$ are:

${x}_{3} = \frac{1}{3} \left(5 + \sqrt[3]{- 119 + 12 \sqrt{114}} + \sqrt[3]{- 119 - 12 \sqrt{114}}\right)$

${x}_{4} = \frac{1}{3} \left(5 + \omega \sqrt[3]{- 119 + 12 \sqrt{114}} + {\omega}^{2} \sqrt[3]{- 119 - 12 \sqrt{114}}\right)$

${x}_{5} = \frac{1}{3} \left(5 + {\omega}^{2} \sqrt[3]{- 119 + 12 \sqrt{114}} + \omega \sqrt[3]{- 119 - 12 \sqrt{114}}\right)$