what are the zeros of f(x)=(x-3)(x^3-1)?
You asked:
Find the zeros of \(f\left( x \right) = \left(x - 3\right) \cdot \left({x}^{3} - 1\right)\).
MathBot Answer:
The real zeros of \(f(x)=\left(x^{3} - 1\right) \left(x - 3\right)\) are: $$\begin{align*}x &= 1\\x &= 3\end{align*}$$
The complex zeros of \(f(x)=\left(x^{3} - 1\right) \left(x - 3\right)\) are: $$\begin{align*}x &= - \frac{1}{2} - \frac{\sqrt{3} i}{2} \approx -0.5 - 0.866025403784439 i\\x &= - \frac{1}{2} + \frac{\sqrt{3} i}{2} \approx -0.5 + 0.866025403784439 i\end{align*}$$