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

1 Answer
sente
Feb 13, 2016

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))

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