# What are the factors for 6y^2 - 5y^3 - 4?

Jun 16, 2015

$6 {y}^{2} - 5 {y}^{3} - 4 = - 5 \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) \left(y - {y}_{3}\right)$

${y}_{1} = \frac{1}{{u}_{1} + {v}_{1}}$

${y}_{2} = \frac{1}{\omega {u}_{1} + {\omega}^{2} {v}_{1}}$

${y}_{3} = \frac{1}{{\omega}^{2} {u}_{1} + \omega {v}_{1}}$

as explained below...

#### Explanation:

Attempt to solve $f \left(y\right) = - 5 {y}^{3} + 6 {y}^{2} - 4 = 0$

First divide through by $- {y}^{3}$ to get:

$5 - \frac{6}{y} + \frac{4}{y} ^ 3 = 0$

Let $x = \frac{1}{y}$

Then $4 {x}^{3} - 6 x + 5 = 0$

Now let $x = u + v$

$0 = 4 {\left(u + v\right)}^{3} - 6 \left(u + v\right) + 5$

$= 4 {u}^{3} + 4 {v}^{3} + \left(12 u v - 6\right) \left(u + v\right) + 5$

$= 4 {u}^{3} + 4 {v}^{3} + 6 \left(2 u v - 1\right) \left(u + v\right) + 5$

Let $v = \frac{1}{2 u}$

$= 4 {u}^{3} + \frac{1}{2 {u}^{3}} + 5$

Multiply through by $2 {u}^{3}$ to get:

$8 {\left({u}^{3}\right)}^{2} + 10 \left({u}^{3}\right) + 1 = 0$

${u}^{3} = \frac{- 10 \pm \sqrt{100 - 32}}{16}$

$= \frac{- 10 \pm \sqrt{68}}{16}$

$= \frac{- 5 \pm \sqrt{17}}{8}$

Write:

${u}_{1} = \sqrt[3]{\frac{- 5 + \sqrt{17}}{8}}$

${v}_{1} = \sqrt[3]{\frac{- 5 - \sqrt{17}}{8}}$

Then real root of $4 {x}^{3} - 6 x + 5 = 0$ is

$x = {u}_{1} + {v}_{1}$

The other two (complex) roots are:

$x = \omega {u}_{1} + {\omega}^{2} {v}_{1}$

$x = {\omega}^{2} {u}_{1} + \omega {v}_{1}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

$y = \frac{1}{x}$

So the real root of $f \left(y\right) = 0$ is ${y}_{1} = \frac{1}{{u}_{1} + {v}_{1}}$

and the complex roots are:

${y}_{2} = \frac{1}{\omega {u}_{1} + {\omega}^{2} {v}_{1}}$

${y}_{3} = \frac{1}{{\omega}^{2} {u}_{1} + \omega {v}_{1}}$

$f \left(y\right) = - 5 \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) \left(y - {y}_{3}\right)$