# How do you use the rational roots theorem to find all possible zeros of f(x) = 2x^3 + 8x^2 + 7x - 8?

Aug 13, 2016

$f \left(x\right)$ has no rational zeros, so use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{6} \left(- 8 + \sqrt[3]{- 424 + 54 \sqrt{58}} + \sqrt[3]{- 424 - 54 \sqrt{58}}\right)$

and related Complex zeros.

#### Explanation:

$f \left(x\right) = 2 {x}^{3} + 8 {x}^{2} + 7 x - 8$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ can be expressed in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 8$ and $q$ a divisor of the coefficient $2$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{2} , \pm 1 , \pm 2 , \pm 4 , \pm 8$

None of these work, so we need other means to find the zeros:

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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, $a = 2$, $b = 8$, $c = 7$ and $d = - 8$, so we find:

$\Delta = 3136 - 2744 + 16384 - 6912 - 16128 = - 6264$

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.

$0 = 108 f \left(x\right) = 216 {x}^{3} + 864 {x}^{2} + 756 x - 864$

$= {\left(6 x + 8\right)}^{3} - 66 \left(6 x + 8\right) - 848$

$= {t}^{3} - 66 t - 848$

where $t = \left(6 x + 8\right)$

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Cardano's method

We want to solve:

${t}^{3} - 66 t - 848 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 22\right) \left(u + v\right) - 848 = 0$

Add the constraint $v = \frac{22}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{10648}{u} ^ 3 - 848 = 0$

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

${\left({u}^{3}\right)}^{2} - 848 \left({u}^{3}\right) + 10648 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{848 \pm \sqrt{{\left(- 848\right)}^{2} - 4 \left(1\right) \left(10648\right)}}{2 \cdot 1}$

$= \frac{- 848 \pm \sqrt{719104 - 42592}}{2}$

$= \frac{- 848 \pm \sqrt{676512}}{2}$

$= \frac{- 848 \pm 108 \sqrt{58}}{2}$

$= - 424 \pm 54 \sqrt{58}$

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:

${t}_{1} = \sqrt[3]{- 424 + 54 \sqrt{58}} + \sqrt[3]{- 424 - 54 \sqrt{58}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{- 424 + 54 \sqrt{58}} + {\omega}^{2} \sqrt[3]{- 424 - 54 \sqrt{58}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{- 424 + 54 \sqrt{58}} + \omega \sqrt[3]{- 424 - 54 \sqrt{58}}$

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

Now $x = \frac{1}{6} \left(- 8 + t\right)$. So the zeros of our original cubic are:

${x}_{1} = \frac{1}{6} \left(- 8 + \sqrt[3]{- 424 + 54 \sqrt{58}} + \sqrt[3]{- 424 - 54 \sqrt{58}}\right)$

${x}_{2} = \frac{1}{6} \left(- 8 + \omega \sqrt[3]{- 424 + 54 \sqrt{58}} + {\omega}^{2} \sqrt[3]{- 424 - 54 \sqrt{58}}\right)$

${x}_{3} = \frac{1}{6} \left(- 8 + {\omega}^{2} \sqrt[3]{- 424 + 54 \sqrt{58}} + \omega \sqrt[3]{- 424 - 54 \sqrt{58}}\right)$