How do you factor #x^3 + 4 = 0#?

1 Answer
Feb 13, 2016

Answer:

#x^3+4=(x+4^(1/3))(x^2-4^(1/3)x+4^(2/3))#

#= (x+4^(1/3))(x-2^(-1/3)(1+sqrt(3)i))(x-2^(-1/3)(1-sqrt(3)i))#

Explanation:

Using the sum of cubes formula #a^3+b^3 = (a+b)(a^2-ab+b^2)# we have:

#x^3+4 = x^3+(4^(1/3))^3#

#=(x+4^(1/3))(x^2-4^(1/3)x+4^(2/3))#

If we wish to restrict ourselves to real numbers, we are done. If we allow for complex numbers, then we may continue using the quadratic formula to factor #x^2-4^(1/3)x+4^(2/3)#

#x = (4^(1/3)+-sqrt(4^(2/3)-4(1)(4^(2/3))))/2#

#=2^(-1/3)(1+-sqrt(-3))#

#=2^(-1/3)(1+-sqrt(3)i)#

Thus

#x^3+4 = (x+4^(1/3))(x-2^(-1/3)(1+sqrt(3)i))(x-2^(-1/3)(1-sqrt(3)i))#