How do you find all the zeros of  x^3-4x^2+3x-1?

Aug 8, 2016

Use Cardano's method to find Real zero:

$\frac{1}{3} \left(4 + \sqrt[3]{\frac{- 47 + 3 \sqrt{93}}{2}} + \sqrt[3]{\frac{- 47 - 3 \sqrt{93}}{2}}\right)$

and Complex conjugate pair of related zeros.

Explanation:

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

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

$\Delta = 144 - 108 - 256 - 27 + 216 = - 31$

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.

$0 = 27 f \left(x\right) = 27 {x}^{3} - 108 {x}^{2} + 81 x - 27$

$= {\left(3 x - 4\right)}^{3} - 21 \left(3 x - 4\right) - 47$

$= {t}^{3} - 21 t - 47$

where $t = \left(3 x - 4\right)$

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Cardano's method

We want to solve:

${t}^{3} - 21 t - 47 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 7\right) \left(u + v\right) - 47 = 0$

Add the constraint $v = \frac{7}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{343}{u} ^ 3 - 47 = 0$

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

${\left({u}^{3}\right)}^{2} - 47 \left({u}^{3}\right) + 343 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{47 \pm \sqrt{{\left(- 47\right)}^{2} - 4 \left(1\right) \left(343\right)}}{2 \cdot 1}$

$= \frac{- 47 \pm \sqrt{2209 - 1372}}{2}$

$= \frac{- 47 \pm \sqrt{837}}{2}$

$= \frac{- 47 \pm 3 \sqrt{93}}{2}$

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:

${t}_{1} = \sqrt[3]{\frac{- 47 + 3 \sqrt{93}}{2}} + \sqrt[3]{\frac{- 47 - 3 \sqrt{93}}{2}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{\frac{- 47 + 3 \sqrt{93}}{2}} + {\omega}^{2} \sqrt[3]{\frac{- 47 - 3 \sqrt{93}}{2}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{\frac{- 47 + 3 \sqrt{93}}{2}} + \omega \sqrt[3]{\frac{- 47 - 3 \sqrt{93}}{2}}$

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

Now $x = \frac{1}{3} \left(4 + t\right)$. So the roots of our original cubic are:

${x}_{1} = \frac{1}{3} \left(4 + \sqrt[3]{\frac{- 47 + 3 \sqrt{93}}{2}} + \sqrt[3]{\frac{- 47 - 3 \sqrt{93}}{2}}\right)$

${x}_{2} = \frac{1}{3} \left(4 + \omega \sqrt[3]{\frac{- 47 + 3 \sqrt{93}}{2}} + {\omega}^{2} \sqrt[3]{\frac{- 47 - 3 \sqrt{93}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left(4 + {\omega}^{2} \sqrt[3]{\frac{- 47 + 3 \sqrt{93}}{2}} + \omega \sqrt[3]{\frac{- 47 - 3 \sqrt{93}}{2}}\right)$