(x-1)(x²+x+1)=0
To solve the equation \((x-1)(x^2+x+1) = 0\), we can use the zero product property, which states that if a product of factors equals zero, at least one of the factors must be zero.
Set the first factor equal to zero: \[ x - 1 = 0 \] Solving for \(x\), we get: \[ x = 1 \]
Set the second factor equal to zero: \[ x^2 + x + 1 = 0 \] To solve this quadratic equation, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = 1\), and \(c = 1\).
First, calculate the discriminant (\(b^2 - 4ac\)): \[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \] Since the discriminant is negative, there are no real roots for this quadratic equation, but there are two complex roots.
Now, substitute into the quadratic formula: \[ x = \frac{-1 \pm \sqrt{-3}}{2 \cdot 1} = \frac{-1 \pm i\sqrt{3}}{2} \] Therefore, the roots from this factor are: \[ x = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad x = \frac{-1 - i\sqrt{3}}{2} \]
Putting it all together, the solutions to the equation \((x-1)(x^2+x+1)=0\) are: \[ x = 1, \quad x = \frac{-1 + i\sqrt{3}}{2}, \quad \text{and} \quad x = \frac{-1 - i\sqrt{3}}{2} \]
