# How do you use the rational roots theorem to find all possible zeros of f(x) = -2x^3 + 3x^2 -4x + 8?

Aug 5, 2016

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

It has one positive Real zero and a Complex conjugate pair of non-Real zeros.

#### Explanation:

$f \left(x\right) = - 2 {x}^{3} + 3 {x}^{2} - 4 x + 8$

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 $8$ and $q$ a divisor of the coefficient $- 2$ of the leading term.

That means that the only possible rational zeros are:

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

Note that $f \left(- x\right) = 2 {x}^{3} + 3 {x}^{2} + 4 x + 8$ has no changes of sign of coefficients. So by Descartes rule of signs, $f \left(x\right)$ has no negative zeros.

So the only possible rational zeros are:

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

When $x = \pm 1$ all of the terms will be even except $3 {x}^{2} = 3$. Hence $f \left(\pm 1\right)$ is odd and non-zero.

When $x = \pm 2$ all of the terms will be divisible by $8$ except $3 {x}^{2} = 12$. Hence $f \left(\pm 2\right) \ne 0$.

When $x = \pm 4$ or $x = \pm 8$ all of the terms will be divisible by $16$ except the constant term $8$. Hence $f \left(\pm 4\right) \ne 0$ and $f \left(\pm 8\right) \ne 0$.

So that only leaves as a possible rational zero:

$\frac{1}{2}$

We find:

$f \left(\frac{1}{2}\right) = - 2 \left(\frac{1}{8}\right) + 3 \left(\frac{1}{4}\right) - 4 \left(\frac{1}{2}\right) + 8$

$= \frac{- 1 + 3 - 8 + 32}{4}$

$= \frac{13}{2}$

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

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Bonus

What can we find out about the zeros of $f \left(x\right)$?

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, $a = - 2$, $b = 3$, $c = - 4$, $d = 8$ and we find:

$\Delta = 144 - 512 - 864 - 6912 + 3456 = - 4688$

Since $\Delta < 0$, this cubic has exactly one Real zero (which we already know is positive) and two non-Real Complex zeros which are Complex conjugates of one another.

Let's simplify $f \left(x\right)$ by linear substitution (called a Tschirnhaus tranformation) as follows:

$0 = - 4 f \left(x\right) = 8 {x}^{3} - 12 {x}^{2} + 16 x - 32$

$= {\left(2 x - 1\right)}^{3} + 5 \left(2 x - 1\right) - 26$

$= {t}^{3} + 5 t - 26$

where $t = 2 x - 1$

Use Cardano's method to find the roots:

Let $t = u + v$

Then we want to solve:

${u}^{3} + {v}^{3} + \left(3 u v + 5\right) \left(u + v\right) - 26 = 0$

Add the constraint $v = - \frac{5}{3 u}$ to eliminate the term in $\left(u + v\right)$ and get:

${u}^{3} - \frac{125}{27 {u}^{3}} - 26 = 0$

Multiply through by $27 {u}^{3}$ and rearrange to get:

$27 {\left({u}^{3}\right)}^{2} - 702 \left({u}^{3}\right) - 125 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{702 \pm \sqrt{{702}^{2} + 4 \left(27\right) \left(125\right)}}{2 \cdot 27}$

$= \frac{702 \pm \sqrt{492804 + 13500}}{54}$

$= \frac{702 \pm \sqrt{506304}}{54}$

$= \frac{702 \pm 24 \sqrt{879}}{54}$

$= \frac{351 \pm 12 \sqrt{879}}{27}$

Since ${u}^{3}$ is 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 the Real root:

${t}_{1} = \frac{1}{3} \left(\sqrt[3]{351 + 12 \sqrt{879}} + \sqrt[3]{351 - 12 \sqrt{879}}\right)$

and related Complex roots:

${t}_{2} = \frac{1}{3} \left(\omega \sqrt[3]{351 + 12 \sqrt{879}} + {\omega}^{2} \sqrt[3]{351 - 12 \sqrt{879}}\right)$

${t}_{3} = \frac{1}{3} \left({\omega}^{2} \sqrt[3]{351 + 12 \sqrt{879}} + \omega \sqrt[3]{351 - 12 \sqrt{879}}\right)$

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

Then $x = \frac{1}{2} \left(1 + t\right)$ hence $f \left(x\right)$ has Real zero:

${x}_{1} = \frac{1}{2} \left(1 + \frac{1}{3} \left(\sqrt[3]{351 + 12 \sqrt{879}} + \sqrt[3]{351 - 12 \sqrt{879}}\right)\right)$

and related Complex zeros:

${x}_{2} = \frac{1}{2} \left(1 + \frac{1}{3} \left(\omega \sqrt[3]{351 + 12 \sqrt{879}} + {\omega}^{2} \sqrt[3]{351 - 12 \sqrt{879}}\right)\right)$

${x}_{3} = \frac{1}{2} \left(1 + \frac{1}{3} \left({\omega}^{2} \sqrt[3]{351 + 12 \sqrt{879}} + \omega \sqrt[3]{351 - 12 \sqrt{879}}\right)\right)$