# What are all the possible rational zeros for f(x)=-6x^3+5x^2-2x+18 and how do you find all zeros?

Jan 20, 2018

$x = \sqrt[3]{\frac{100345392 + \sqrt{{100345392}^{2} + 181081080576}}{68024448}} + \sqrt[3]{\frac{100345392 - \sqrt{{100345392}^{2} + 181081080576}}{68024448}} + \frac{5}{18}$

#### Explanation:

We can use the rational roots theorem to find that all the possible rational roots are $\pm$ all of the factors of $18$ divided by all of the factors of $6$:
$\pm 18 , \pm 9 , \pm 6 , \pm \frac{9}{2} , \pm 3 , \pm \frac{3}{2} , \pm 1 , \pm \frac{2}{3} , \pm \frac{1}{2} , \pm \frac{1}{3} , \pm \frac{1}{6}$

By using Descartes' rule of signs and showing that $f \left(- x\right)$ has no sign changes, we can exclude all the negative possibilities:
$18 , 9 , 6 , \frac{9}{2} , 3 , \frac{3}{2} , 1 , \frac{2}{3} , \frac{1}{2} , \frac{1}{3} , \frac{1}{6}$

After trying these, we find that none of them are zeroes. This means that the polynomial has no rational zeroes.

We can also compute the cubic discriminant and get a value of −304580. This means that the polynomial has one real solution and two complex ones.

To find the real solution, I will first transform the equation into a depressed cubic. This is done by first dividing so the leading coefficient is $1$:
${x}^{3} - \frac{5}{6} {x}^{2} + \frac{1}{3} x - 3 = 0$

Then we substitute $x = t + \frac{5}{18}$:
${\left(t + \frac{5}{18}\right)}^{3} - \frac{5}{6} {\left(t + \frac{5}{18}\right)}^{2} + \frac{1}{3} \left(t + \frac{5}{18}\right) - 3 = 0$

${t}^{3} + \frac{11}{108} t - \frac{8603}{2916} = 0$

Now that we have a depressed cubic (ie no ${t}^{2}$'s), we can solve by substituting $t = u + v$:
${\left(u + v\right)}^{3} + \frac{11}{108} \left(u + v\right) - \frac{8603}{2916} = 0$

${u}^{3} + {v}^{3} + 3 {u}^{2} v + 3 u {v}^{2} + \frac{11}{108} u + \frac{11}{108} v - \frac{8603}{2916} = 0$

Now we can factor:
${u}^{3} + {v}^{3} + 3 u v \left(u + v\right) + \frac{11}{108} \left(u + v\right) - \frac{8603}{2916} = 0$

${u}^{3} + {v}^{3} + \left(3 u v + \frac{11}{108}\right) \left(u + v\right) - \frac{8603}{2916} = 0$

Since $u$ and $v$ are any arbitrary numbers, I can add a condition that $3 u v + \frac{11}{108} = 0$. This allows us to get rid of the middle term and rewrite $v$ in terms of $u$:
${u}^{3} + {\left(- \frac{11}{324 u}\right)}^{3} - \frac{8603}{2916} = 0$

${u}^{3} - \frac{1331}{34012224 {u}^{3}} - \frac{8603}{2916} = 0$

Multiply through by ${u}^{3}$:
${u}^{6} - \frac{8603}{2916} {u}^{3} - \frac{1331}{34012224} = 0$

This is a quadratic in ${u}^{3}$, so we can solve using the quadratic formula:
${u}^{3} = \frac{100345392 \pm \sqrt{{100345392}^{2} + 181081080576}}{68024448}$

$u = \sqrt[3]{\frac{100345392 \pm \sqrt{{100345392}^{2} + 181081080576}}{68024448}}$

We can interpret one of the roots as the value for $u$, and the other for $v$. This gives that $t$ is equal to:
$t = u + v = \sqrt[3]{\frac{100345392 + \sqrt{{100345392}^{2} + 181081080576}}{68024448}} + \sqrt[3]{\frac{100345392 - \sqrt{{100345392}^{2} + 181081080576}}{68024448}}$

Now, we just put back the solution for $t$ into $x = t + \frac{5}{18}$ to get the solution for $x$:
$x = \sqrt[3]{\frac{100345392 + \sqrt{{100345392}^{2} + 181081080576}}{68024448}} + \sqrt[3]{\frac{100345392 - \sqrt{{100345392}^{2} + 181081080576}}{68024448}} + \frac{5}{18}$