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

Aug 11, 2016

Find $f \left(x\right)$ has no rational zeros.

Use Cardano's method to find Real zero:

${x}_{1} = \frac{1}{6} \left(- 7 + \sqrt[3]{- 593 + 12 \sqrt{2442}} + \sqrt[3]{- 593 - 12 \sqrt{2442}}\right)$

and related Complex zeros.

#### Explanation:

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

By the rational roots 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 $- 8$ and $q$ a divisor of the coefficient $2$ of the leading term. So the only possible rational zeros are:

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

None of these works, so $f \left(x\right)$ has no rational zeros.

We can find the irrational zeros using Cardano's method.

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

$\Delta = 3136 - 4096 + 10976 - 6912 - 16128 = - 13024$

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.

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

$= {\left(6 x + 7\right)}^{3} - 3 \left(6 x + 7\right) - 1186$

$= {t}^{3} - 3 t - 1186$

where $t = \left(6 x + 7\right)$

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Cardano's method

We want to solve:

${t}^{3} - 3 t - 1186 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 1\right) \left(u + v\right) - 1186 = 0$

Add the constraint $v = \frac{1}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{1}{u} ^ 3 - 1186 = 0$

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

${\left({u}^{3}\right)}^{2} - 1186 \left({u}^{3}\right) + 1 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{1186 \pm \sqrt{{\left(- 1186\right)}^{2} - 4 \left(1\right) \left(1\right)}}{2 \cdot 1}$

$= \frac{- 1186 \pm \sqrt{1406596 - 4}}{2}$

$= \frac{- 1186 \pm \sqrt{1406592}}{2}$

$= \frac{- 1186 \pm 24 \sqrt{2442}}{2}$

$= - 593 \pm 12 \sqrt{2442}$

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=root(3)(-593+12sqrt(2442))+root(3)((-593-12sqrt(2442))

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{- 593 + 12 \sqrt{2442}} + {\omega}^{2} \sqrt[3]{- 593 - 12 \sqrt{2442}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{- 593 + 12 \sqrt{2442}} + \omega \sqrt[3]{- 593 - 12 \sqrt{2442}}$

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

Now $x = \frac{1}{6} \left(- 7 + t\right)$. So the zeros of our original cubic are:

${x}_{1} = \frac{1}{6} \left(- 7 + \sqrt[3]{- 593 + 12 \sqrt{2442}} + \sqrt[3]{- 593 - 12 \sqrt{2442}}\right)$

${x}_{2} = \frac{1}{6} \left(- 7 + \omega \sqrt[3]{- 593 + 12 \sqrt{2442}} + {\omega}^{2} \sqrt[3]{- 593 - 12 \sqrt{2442}}\right)$

${x}_{3} = \frac{1}{6} \left(- 7 + {\omega}^{2} \sqrt[3]{- 593 + 12 \sqrt{2442}} + \omega \sqrt[3]{- 593 - 12 \sqrt{2442}}\right)$