To solve the equation ((x + 2)(x^2 - 2x + 4) = 7), we can start by expanding the left side:
- Expand the expression ((x + 2)(x^2 - 2x + 4)):
[
(x + 2)(x^2 - 2x + 4) = x(x^2 - 2x + 4) + 2(x^2 - 2x + 4)
]
[
= x^3 - 2x^2 + 4x + 2x^2 - 4x + 8
]
[
= x^3 + 8
]
- Now set the equation equal to 7:
[
x^3 + 8 = 7
]
- Subtract 7 from both sides:
[
x^3 + 8 - 7 = 0
]
[
x^3 + 1 = 0
]
- This can be factored as:
[
(x + 1)(x^2 - x + 1) = 0
]
- Now, set each factor equal to zero:
For (x + 1 = 0):
[
x = -1
]
For (x^2 - x + 1 = 0), we use the quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}):
Here, (a = 1), (b = -1), and (c = 1).
Calculate the discriminant:
[
b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3
]
Since the discriminant is negative, the quadratic equation has no real roots. The roots are complex and are given as:
[
x = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}
]
- Thus, the solutions to the original equation ((x + 2)(x^2 - 2x + 4) = 7) are:
- (x = -1)
- (x = \frac{1 + i\sqrt{3}}{2})
- (x = \frac{1 - i\sqrt{3}}{2})
So the complete set of solutions is:
[
\boxed{-1, \frac{1 + i\sqrt{3}}{2}, \frac{1 - i\sqrt{3}}{2}}
]
