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

1 Answer
Feb 24, 2016

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)))