Posted by Martha on Friday, October 5, 2012 at 1:14am.
(x+2)(x^2-2x+4)(x-2)(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
(x-1)^2 = -3
x-1 = ±√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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