# How do you find all the zeros of f(x)=x^4-x^3-20x^2+10x+27?

Aug 13, 2016

Long description of how to solve this (typical) quartic algebraically...

#### Explanation:

$f \left(x\right) = {x}^{4} - {x}^{3} - 20 {x}^{2} + 10 x + 27$

This is a typical quartic, which is algebraically solvable, but gets very messy. It is possible to find approximations to the zeros numerically, but what do you learn from that?

Let's see at least the direction of how to solve this quartic algebraically...

Tschirnhaus transformation

First we simplify the quartic using a linear substitution called a Tschirnhaus transformation.

$256 f \left(x\right) = 256 {x}^{4} - 256 {x}^{3} - 5120 {x}^{2} + 2560 x + 6912$

$= {\left(4 x - 1\right)}^{4} - 326 {\left(4 x - 1\right)}^{2} - 8 \left(4 x - 1\right) + 7229$

$= {t}^{4} - 326 {t}^{2} - 8 t + 7229$

where $t = \left(4 x - 1\right)$.

Since this quartic in $t$ is monic and has no ${t}^{3}$ term, it factors as a product of two monic quadratics with opposite middle terms:

${t}^{4} - 326 {t}^{2} - 8 t + 7229$

$= \left({t}^{2} - a t + b\right) \left({t}^{2} + a t + c\right)$

$= {t}^{4} + \left(b + c - {a}^{2}\right) {t}^{2} + \left(b - c\right) a t + b c$

Equating coefficients and rearranging slightly we get:

$\left\{\begin{matrix}b + c = {a}^{2} - 326 \\ b - c = - \frac{8}{a} \\ b c = 7229\end{matrix}\right.$

Then:

${\left({a}^{2} - 326\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = \frac{64}{a} ^ 2 + 28916$

Expanding, this becomes:

${\left({a}^{2}\right)}^{2} - 652 \left({a}^{2}\right) + 106276 = \frac{64}{a} ^ 2 + 28916$

Multiply through by ${a}^{2}$ and rearrange slightly to get:

${\left({a}^{2}\right)}^{3} - 652 {\left({a}^{2}\right)}^{2} + 77360 \left({a}^{2}\right) - 64 = 0$

This is a cubic in ${a}^{2}$ which we can attempt to solve. First let's look at its discriminant...

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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, $a = 1$, $b = - 652$, $c = 77360$ and $d = - 64$, so we find:

$\Delta = 726994944 + 126959616 + 2059485184 - 110592 - 2024136704 = 889192448$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

Since all three zeros are Real, methods like Cardano's method which express zeros in terms of $n$th roots will include irreducible cube roots of Complex numbers.

My preferred method for such cubics is trigonometric substitution, but first we want another Tschirnhaus transformation...

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Tschirnhaus transformation

$27 \left({\left({a}^{2}\right)}^{3} - 652 {\left({a}^{2}\right)}^{2} + 77360 \left({a}^{2}\right) - 64\right)$

$= 27 {\left({a}^{2}\right)}^{3} - 17604 {\left({a}^{2}\right)}^{2} + 2088720 \left({a}^{2}\right) - 1728$

$= {\left(3 {a}^{2} - 652\right)}^{3} - 579072 \left(3 {a}^{2} - 652\right) - 100388864$

$= {s}^{3} - 579072 s - 100388864$

where $s = 3 {a}^{2} - 652$

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Trigonometric solution

Let $s = k \cos \theta$ where $k$ is chosen to squeeze the resulting expression into a form including $4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$...

Let $k = 32 \sqrt{754}$

Then:

$0 = {s}^{3} - 579072 s - 100388864$

$= {\left(32 \sqrt{754} \cos \theta\right)}^{3} - 579072 \left(32 \sqrt{754} \cos \theta\right) - 100388864$

$= 6176768 \sqrt{754} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 100388864$

$= 6176768 \sqrt{754} \left(\cos 3 \theta\right) - 100388864$

$= 4096 \left(1508 \sqrt{754} \left(\cos 3 \theta\right) - 24509\right)$

Hence:

${s}_{k} = 32 \sqrt{754} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{24509}{1508 \sqrt{754}}\right) + \frac{2 k \pi}{3}\right) \text{ }$ for $k = 0 , 1 , 2$

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To cut a very long story a little short, we can choose any one of these roots to derive $a = \sqrt{\frac{1}{3} \left(s + 652\right)}$ and hence find:

$b = \frac{1}{2} \left({a}^{2} - 326 - \frac{8}{a}\right)$

$c = \frac{1}{2} \left({a}^{2} - 326 + \frac{8}{a}\right)$

leaving us with two quadratics to solve.