# How do you use the rational roots theorem to find all possible zeros of f(x) = x^3 - 10x^2 + 9x - 24?

Aug 13, 2016

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

${x}_{1} = \frac{1}{3} \left(10 + \sqrt[3]{- 919 + 18 \sqrt{1406}} + \sqrt[3]{- 919 - 18 \sqrt{1406}}\right)$

and related Complex zeros.

#### Explanation:

$f \left(x\right) = {x}^{3} - 10 {x}^{2} + 9 x - 24$

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 $- 24$ and $q$ a divisor of the coefficient $1$ of the leading term. That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 8 , \pm 12 , \pm 24$

In addition, using Descartes' rule of signs we find that $f \left(x\right)$ has $1$ or $3$ positive Real zeros and no negative Real zeros. So the only possible rational zeros are:

$1 , 2 , 3 , 4 , 6 , 8 , 12 , 24$

None of these work, so $f \left(x\right)$ has no rational 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, $a = 1$, $b = - 10$, $c = 9$ and $d = - 24$, so we find:

$\Delta = 8100 - 2916 - 96000 - 15552 + 38880 = - 67488$

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.

$0 = 27 f \left(x\right) = 27 {x}^{3} - 270 {x}^{2} + 243 x - 648$

$= {\left(3 x - 10\right)}^{3} - 219 \left(3 x - 10\right) - 1838$

$= {t}^{3} - 219 t - 1838$

where $t = \left(3 x - 10\right)$

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Cardano's method

We want to solve:

${t}^{3} - 219 t - 1838 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 73\right) \left(u + v\right) - 1838 = 0$

Add the constraint $v = \frac{73}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{389017}{u} ^ 3 - 1838 = 0$

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

${\left({u}^{3}\right)}^{2} - 1838 \left({u}^{3}\right) + 389017 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{1838 \pm \sqrt{{\left(- 1838\right)}^{2} - 4 \left(1\right) \left(389017\right)}}{2 \cdot 1}$

$= \frac{- 1838 \pm \sqrt{3378244 - 1556068}}{2}$

$= \frac{- 1838 \pm \sqrt{1822176}}{2}$

$= \frac{- 1838 \pm 36 \sqrt{1406}}{2}$

$= - 919 \pm 18 \sqrt{1406}$

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:

${t}_{1} = \sqrt[3]{- 919 + 18 \sqrt{1406}} + \sqrt[3]{- 919 - 18 \sqrt{1406}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{- 919 + 18 \sqrt{1406}} + {\omega}^{2} \sqrt[3]{- 919 - 18 \sqrt{1406}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{- 919 + 18 \sqrt{1406}} + \omega \sqrt[3]{- 919 - 18 \sqrt{1406}}$

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

Now $x = \frac{1}{3} \left(10 + t\right)$. So the zeros of our original cubic are:

${x}_{1} = \frac{1}{3} \left(10 + \sqrt[3]{- 919 + 18 \sqrt{1406}} + \sqrt[3]{- 919 - 18 \sqrt{1406}}\right)$

${x}_{2} = \frac{1}{3} \left(10 + \omega \sqrt[3]{- 919 + 18 \sqrt{1406}} + {\omega}^{2} \sqrt[3]{- 919 - 18 \sqrt{1406}}\right)$

${x}_{3} = \frac{1}{3} \left(10 + {\omega}^{2} \sqrt[3]{- 919 + 18 \sqrt{1406}} + \omega \sqrt[3]{- 919 - 18 \sqrt{1406}}\right)$