# What are the roots of 38r^3-27r^2-27r-27 = 0 ?

Feb 5, 2017

Real root:

$x = \frac{3}{2}$

Complex roots:

$x = - \frac{15}{38} \pm \frac{3 \sqrt{51}}{38} i$

#### Explanation:

$f \left(x\right) = 38 {r}^{3} - 27 {r}^{2} - 27 r - 27$

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 $- 27$ and $q$ a divisor of the coefficient $38$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{38} , \pm \frac{1}{19} , \pm \frac{3}{38} , \pm \frac{3}{19} , \pm \frac{9}{38} , \pm \frac{1}{2} , \pm \frac{27}{38} , \pm 1 , \pm \frac{27}{19} , \pm \frac{3}{2} , \pm 3 , \pm \frac{9}{2} , \pm 9 , \pm \frac{27}{2} , \pm 27$

That's rather a lot of possibilities to try, so let's see how we can narrow down the search...

First note that the pattern of signs of the coefficients of $f \left(x\right)$ is $+ - - -$. By Descartes' Rule of Signs, since this has one change, we can deduce that $f \left(x\right)$ has exactly one positive real zero.

We can observe that:

$f \left(1\right) = 38 - 27 - 27 - 27 = - 43 < 0$

So the positive Real zero is greater than $1$.

We also find:

$f \left(3\right) = 38 \left(27\right) - 27 \left(9\right) - 27 \left(3\right) - 27 = 27 \left(38 - 9 - 3 - 1\right) = 27 \cdot 25$

$= 675 > 0$

So let's try:

$f \left(\frac{3}{2}\right) = 38 \left(\frac{27}{8}\right) - 27 \left(\frac{9}{4}\right) - 27 \left(\frac{3}{2}\right) - 27 = 27 \left(\frac{38}{8} - \frac{9}{4} - \frac{3}{2} - 1\right)$

$= 27 \frac{19 - 9 - 6 - 4}{4} = 0$

So $x = \frac{3}{2}$ is a zero and $\left(2 x - 3\right)$ a factor:

$38 {r}^{3} - 27 {r}^{2} - 27 r - 27 = \left(2 x - 3\right) \left(19 {x}^{2} + 15 x + 9\right)$

We can find the zeros of the remaining quadratic using the quadratic formula with $a = 19$, $b = 15$, $c = 9$...

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- 15 \pm \sqrt{{15}^{2} - 4 \left(19\right) \left(9\right)}}{2 \cdot 19}$

$\textcolor{w h i t e}{x} = \frac{- 15 \pm \sqrt{225 - 684}}{38}$

$\textcolor{w h i t e}{x} = \frac{- 15 \pm 3 \sqrt{51} i}{38}$

$\textcolor{w h i t e}{x} = - \frac{15}{38} \pm \frac{3 \sqrt{51}}{38} i$