How do you find all the real and complex roots of #x^3 + 8 = 0#?

1 Answer
Jan 12, 2016

Answer:

Roots: #x in {-2, color(white)("X")1+sqrt(3)i, color(white)("X")1-sqrt(3)i)}#

Explanation:

Since #(-2)^3=-8#
#color(white)("XXX")x=(-2)# is an obvious root of #(x^3+8=0)#

Therefore #(x+2)# is a factor of #(x^3+8)#

#(x^3+8) div (x+2) = x^2-2x+4#
#color(white)("XXX")#use synthetic or polynomial long division to get
#color(white)("XXX")#this result if it is not apparent.

Use the quadratic formula:
#color(white)("XXX")x=(-b+-sqrt(b^2+4ac))/(2a)#
with #a=1, b=(-2), c=4#
to get the remaining (complex) roots
#color(white)("XXX")x=1+-sqrt(3)i#