Question 426949
{{{f(x) = x^4-2x^3-8x^2+8x+16 = x^4-2x^3-4x^2 - 4x^2+8x+16

= x^2(x^2 - 2x - 4) - 4(x^2 - 2x - 4)

=(x^2 - 4)(x^2 - 2x - 4)

= (x - 2)(x+2)(x^2 - 2x - 4) = 0}}}

The real zeroes of {{{x^2 - 2x - 4 }}} are {{{x = (2 +- sqrt( 4-4*1*-4 ))/2 = (2 +- sqrt(20))/2  = (2 +- 2sqrt(5))/2 = (1 +- sqrt(5))}}}.

The other two real zeroes are 2, -2.