# How do you use the rational roots theorem to find all possible zeros of F(X) = 6x^4 + 2x^3 - 6x^2 + 3x - 5 ?

Aug 13, 2016

$F \left(x\right)$ has rational zero $x = 1$, another Real zero:

${x}_{1} = \frac{1}{18} \left(- 8 + \sqrt[3]{2510 + 54 \sqrt{2153}} + \sqrt[3]{2510 - 54 \sqrt{2153}}\right)$

and two related Complex zeros.

#### Explanation:

$F \left(x\right) = 6 {x}^{4} + 2 {x}^{3} - 6 {x}^{2} + 3 x - 5$

By the rational root theorem, any rational zeros of $F \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 5$ and $q$ a divisor of the coefficient $6$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{6} , + \frac{1}{3} , \pm \frac{1}{2} , \pm \frac{5}{6} , \pm 1 , \pm \frac{5}{3} , \pm \frac{5}{2} , \pm 5$

Notice that the sum of the coefficients of $F \left(x\right)$ is zero, hence:

$F \left(1\right) = 6 + 2 - 6 + 3 - 5 = 0$

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

$6 {x}^{4} + 2 {x}^{3} - 6 {x}^{2} + 3 x - 5$

$= \left(x - 1\right) \left(6 {x}^{3} + 8 {x}^{2} + 2 x + 5\right)$

None of the remaining possible rational zeros work, so let's focus on the cubic:

$f \left(x\right) = 6 {x}^{3} + 8 {x}^{2} + 2 x + 5$

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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, $a = 6$, $b = 8$, $c = 2$ and $d = 5$, so we find:

$\Delta = 256 - 192 - 10240 - 24300 + 8640 = - 25836$

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.

$0 = 972 f \left(x\right) = 5832 {x}^{3} + 7776 {x}^{2} + 1944 x + 4860$

$= {\left(18 x + 8\right)}^{3} - 84 \left(18 x + 8\right) + 5020$

$= {t}^{3} - 84 t + 5020$

where $t = \left(18 x + 8\right)$

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Cardano's method

We want to solve:

${t}^{3} - 84 t + 5020 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 28\right) \left(u + v\right) + 5020 = 0$

Add the constraint $v = \frac{28}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{21952}{u} ^ 3 + 5020 = 0$

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

${\left({u}^{3}\right)}^{2} + 5020 \left({u}^{3}\right) + 21952 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{- 5020 \pm \sqrt{{\left(5020\right)}^{2} - 4 \left(1\right) \left(21952\right)}}{2 \cdot 1}$

$= \frac{5020 \pm \sqrt{25200400 - 87808}}{2}$

$= \frac{5020 \pm \sqrt{25112592}}{2}$

$= \frac{5020 \pm 108 \sqrt{2153}}{2}$

$= 2510 \pm 54 \sqrt{2153}$

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:

${t}_{1} = \sqrt[3]{2510 + 54 \sqrt{2153}} + \sqrt[3]{2510 - 54 \sqrt{2153}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{2510 + 54 \sqrt{2153}} + {\omega}^{2} \sqrt[3]{2510 - 54 \sqrt{2153}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{2510 + 54 \sqrt{2153}} + \omega \sqrt[3]{2510 - 54 \sqrt{2153}}$

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

Now $x = \frac{1}{18} \left(- 8 + t\right)$. So the roots of our original cubic are:

${x}_{1} = \frac{1}{18} \left(- 8 + \sqrt[3]{2510 + 54 \sqrt{2153}} + \sqrt[3]{2510 - 54 \sqrt{2153}}\right)$

${x}_{2} = \frac{1}{18} \left(- 8 + \omega \sqrt[3]{2510 + 54 \sqrt{2153}} + {\omega}^{2} \sqrt[3]{2510 - 54 \sqrt{2153}}\right)$

${x}_{3} = \frac{1}{18} \left(- 8 + {\omega}^{2} \sqrt[3]{2510 + 54 \sqrt{2153}} + \omega \sqrt[3]{2510 - 54 \sqrt{2153}}\right)$