college algebra
posted by Martha .
solve x^6  64 = 0

(x+2)(x^22x+4)(x2)(x^2+2x+4)

x ^ 6  64 = 0 Add 64 to both sides
x ^ 6  64 + 64 = 0 + 64
x ^ 6 = 64
x = sixth root of 64
x = + OR  2
Solutions :
x =  2
and
x = 2 
since you have a x^6 .... there will be 6 solutions including complex numbers.
Bosnian gave you the two real roots,
Monic only factored the expression, using the factored form we can find the 4 complex roots
x^2  2x + 4 = 0
x^2  2x + 1 = 4+1
(x1)^2 = 3
x1 = ±√3 i
x = 1 ± √3 i
In a similar way
x^2 + 2x + 4 = 0 gives us
x = 1 ±√3 i
(when solving quadratics with a coefficient of 1 for the x^2 term and an EVEN middle coefficient, I always use completing the square rather than the formula, it is faster and yields the answer in reduced form) 
x+8<4 help me self this please,

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