Question

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

Answer 1

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

`8x^4-6x^3-83x^2-6x+8=0`

Explanation:

Let `f(x) = 8x^4-6x^3-83x^2-6x+8=0`

Due to the symmetry of the coefficients, if `r` is a root then so is `1/r`...

`r^4 f(1/r) = r^4 (8r^(-4)-6r^(-3)-83r^(-2)-6r^(-1)+8)`

`=8-6r-83r^2-6r^3+8r^4 = f(r)`

Hence `f(x)` must have a factorisation of the form:

`f(x) = 8(x^2+ax+1)(x^2+bx+1)`

`=8(x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1)`

`=8x^4+8(a+b)x^3+8(2+ab)x^2+8(a+b)x+8`

Equating coefficients, we find:

`8(a+b) = -6`

`8(2+ab) = -83`

Hence:

`a+b = -3/4`

`ab = -83/8-2 = -99/8`

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

`t^2+3/4t-99/8 = 0`

That is:

`8t^2+6t-99 = 0`

Use the quadratic formula to find roots:

`(-6+-sqrt(6^2-(4*8*-99)))/(2*8)`

`=(-6+-sqrt(36+3168))/16`

`=(-6+-sqrt(3204))/16`

`=(-6+-6 sqrt(89))/16`

`=(-3+-3 sqrt(89))/8`

Hence:

`f(x) = 8(x^2+3/8(1+sqrt(89))x+1)(x^2+3/8(1-sqrt(89))+1)`

`=1/8(8x^2+3(1+sqrt(89))x+8)(8x^2+3(1-sqrt(89))x+8)`

Hence using the quadratic formula twice more:

`x = (-3(1+sqrt(89))+-sqrt(9(1+sqrt(89))^2-256))/16`

`=1/16(-3(1+sqrt(89))+-sqrt(9(90+2sqrt(89))-256))`

`=1/16(-3(1+sqrt(89))+-sqrt(554+18sqrt(89)))`

`x = (-3(1-sqrt(89))+-sqrt(9(1-sqrt(89))^2-256))/16`

`=1/16(-3(1-sqrt(89))+-sqrt(9(90-2sqrt(89))-256))`

`=1/16(-3(1-sqrt(89))+-sqrt(554-18sqrt(89)))`