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

Answer:

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)))#