# How do you find all the zeros of 4x^3-2x^2-x+6?

Jul 6, 2016

Use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{6} \left(1 + \sqrt[3]{\frac{- 313 + 3 \sqrt{10857}}{2}} + \sqrt[3]{\frac{- 313 - 3 \sqrt{10857}}{2}}\right)$

and related Complex zeros.

#### Explanation:

$f \left(x\right) = 4 {x}^{3} - 2 {x}^{2} - x + 6$

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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, $a = 4$, $b = - 2$, $c = - 1$ and $d = 6$, so we find:

$\Delta = 4 + 16 + 192 - 15552 + 864 = - 14476$

Since $\Delta < 0$ this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

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Tschirnhaus transformation

To simplify the problem, eliminate any square term using a linear substitution. This is a simple form of what is called a Tschirnhaus transformation. To reduce the amount of arithmetic involving fractions, first multiply by $2 \cdot {3}^{3} = 54$ ...

$54 f \left(x\right) = 216 {x}^{3} - 108 {x}^{2} - 54 x + 324$

$= {\left(6 x - 1\right)}^{3} - 12 \left(6 x - 1\right) + 313$

$= {t}^{3} - 12 t + 313$

where $t = 6 x - 1$

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Cardano's method

To solve ${t}^{3} - 12 t + 313 = 0$, use the substitution $t = u + v$ to get:

${u}^{3} + {v}^{3} + 3 \left(u v - 4\right) \left(u + v\right) + 313 = 0$

To eliminate the term in $\left(u + v\right)$ add the constraint $v = \frac{4}{u}$ to get:

${u}^{3} + \frac{64}{u} ^ 3 + 313 = 0$

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

${\left({u}^{3}\right)}^{2} + 313 \left({u}^{3}\right) + 64 = 0$

Using the quadratic formula, we find:

${u}^{3} = \frac{- 313 \pm \sqrt{{313}^{2} - \left(4 \cdot 1 \cdot 64\right)}}{2 \cdot 1}$

$= \frac{- 313 \pm \sqrt{97969 - 256}}{2}$

$= \frac{- 313 \pm \sqrt{97713}}{2}$

$= \frac{- 313 \pm 3 \sqrt{10857}}{2}$

Since these roots are Real and the derivation was 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:

${t}_{1} = \sqrt[3]{\frac{- 313 + 3 \sqrt{10857}}{2}} + \sqrt[3]{\frac{- 313 - 3 \sqrt{10857}}{2}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{\frac{- 313 + 3 \sqrt{10857}}{2}} + {\omega}^{2} \sqrt[3]{\frac{- 313 - 3 \sqrt{10857}}{2}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{\frac{- 313 + 3 \sqrt{10857}}{2}} + \omega \sqrt[3]{\frac{- 313 - 3 \sqrt{10857}}{2}}$

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

Then $x = \frac{1}{6} \left(1 + t\right)$, hence zeros of our original cubic:

${x}_{1} = \frac{1}{6} \left(1 + \sqrt[3]{\frac{- 313 + 3 \sqrt{10857}}{2}} + \sqrt[3]{\frac{- 313 - 3 \sqrt{10857}}{2}}\right)$

${x}_{2} = \frac{1}{6} \left(1 + \omega \sqrt[3]{\frac{- 313 + 3 \sqrt{10857}}{2}} + {\omega}^{2} \sqrt[3]{\frac{- 313 - 3 \sqrt{10857}}{2}}\right)$

${x}_{3} = \frac{1}{6} \left(1 + {\omega}^{2} \sqrt[3]{\frac{- 313 + 3 \sqrt{10857}}{2}} + \omega \sqrt[3]{\frac{- 313 - 3 \sqrt{10857}}{2}}\right)$