# How do you find all the real and complex roots of 8x^5-6x^4-83x^2-6x+8=0?

Feb 24, 2016

Here's an answer based on the supposition that the equation should have been:

$8 {x}^{4} - 6 {x}^{3} - 83 {x}^{2} - 6 x + 8 = 0$

#### Explanation:

Let $f \left(x\right) = 8 {x}^{4} - 6 {x}^{3} - 83 {x}^{2} - 6 x + 8 = 0$

Due to the symmetry of the coefficients, if $r$ is a root then so is $\frac{1}{r}$...

${r}^{4} f \left(\frac{1}{r}\right) = {r}^{4} \left(8 {r}^{- 4} - 6 {r}^{- 3} - 83 {r}^{- 2} - 6 {r}^{- 1} + 8\right)$

$= 8 - 6 r - 83 {r}^{2} - 6 {r}^{3} + 8 {r}^{4} = f \left(r\right)$

Hence $f \left(x\right)$ must have a factorisation of the form:

$f \left(x\right) = 8 \left({x}^{2} + a x + 1\right) \left({x}^{2} + b x + 1\right)$

$= 8 \left({x}^{4} + \left(a + b\right) {x}^{3} + \left(2 + a b\right) {x}^{2} + \left(a + b\right) x + 1\right)$

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

Equating coefficients, we find:

$8 \left(a + b\right) = - 6$

$8 \left(2 + a b\right) = - 83$

Hence:

$a + b = - \frac{3}{4}$

$a b = - \frac{83}{8} - 2 = - \frac{99}{8}$

So $a$ and $b$ are the roots of:

${t}^{2} + \frac{3}{4} t - \frac{99}{8} = 0$

That is:

$8 {t}^{2} + 6 t - 99 = 0$

Use the quadratic formula to find roots:

$\frac{- 6 \pm \sqrt{{6}^{2} - \left(4 \cdot 8 \cdot - 99\right)}}{2 \cdot 8}$

$= \frac{- 6 \pm \sqrt{36 + 3168}}{16}$

$= \frac{- 6 \pm \sqrt{3204}}{16}$

$= \frac{- 6 \pm 6 \sqrt{89}}{16}$

$= \frac{- 3 \pm 3 \sqrt{89}}{8}$

Hence:

$f \left(x\right) = 8 \left({x}^{2} + \frac{3}{8} \left(1 + \sqrt{89}\right) x + 1\right) \left({x}^{2} + \frac{3}{8} \left(1 - \sqrt{89}\right) + 1\right)$

$= \frac{1}{8} \left(8 {x}^{2} + 3 \left(1 + \sqrt{89}\right) x + 8\right) \left(8 {x}^{2} + 3 \left(1 - \sqrt{89}\right) x + 8\right)$

Hence using the quadratic formula twice more:

$x = \frac{- 3 \left(1 + \sqrt{89}\right) \pm \sqrt{9 {\left(1 + \sqrt{89}\right)}^{2} - 256}}{16}$

$= \frac{1}{16} \left(- 3 \left(1 + \sqrt{89}\right) \pm \sqrt{9 \left(90 + 2 \sqrt{89}\right) - 256}\right)$

$= \frac{1}{16} \left(- 3 \left(1 + \sqrt{89}\right) \pm \sqrt{554 + 18 \sqrt{89}}\right)$

$x = \frac{- 3 \left(1 - \sqrt{89}\right) \pm \sqrt{9 {\left(1 - \sqrt{89}\right)}^{2} - 256}}{16}$

$= \frac{1}{16} \left(- 3 \left(1 - \sqrt{89}\right) \pm \sqrt{9 \left(90 - 2 \sqrt{89}\right) - 256}\right)$

$= \frac{1}{16} \left(- 3 \left(1 - \sqrt{89}\right) \pm \sqrt{554 - 18 \sqrt{89}}\right)$