To find the solutions of the polynomial equation x³ + 8 = 0, we start by rewriting the equation in a simpler form:
x³ = -8
Next, we can take the cube root of both sides to find the real solution. The cube root of -8 is:
x = ∛(-8) = -2
So, x = -2 is our real solution. However, we also need to find the complex solutions. The equation can be factored using the identity for the sum of cubes:
a³ + b³ = (a + b)(a² – ab + b²)
Here, we can set a = x and b = 2, which gives us:
(x + 2)(x² – 2x + 4) = 0
This tells us that one solution is x = -2. Now, we need to find the roots of the quadratic x² – 2x + 4 = 0. To find the roots, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / 2a where a = 1, b = -2, c = 4.
Calculating the discriminant:
Δ = (-2)² – 4(1)(4) = 4 – 16 = -12
Since the discriminant is negative, we have complex solutions:
x = (2 ± √(-12)) / 2 = 1 ± i√3
So, the complex solutions are:
x = 1 + i√3 and x = 1 – i√3.
To summarize, the polynomial equation x³ + 8 = 0 has one real solution:
- x = -2
And two complex solutions:
- x = 1 + i√3
- x = 1 – i√3