Find all real and imaginary solutions to the equation. \[ n^{3}-4 n^{2}-n+4=0 \]

To find the solutions to the equation \( n^{3}-4 n^{2}-n+4=0 \), we can start by using the Rational Root Theorem to test for potential rational roots among the factors of the constant term (4) and the leading coefficient (1). The possible rational roots are: \( \pm 1, \pm 2, \pm 4 \). Testing \( n = 1 \): \[ 1^3 - 4(1^2) - 1 + 4 = 1 - 4 - 1 + 4 = 0 \] Since \( n = 1 \) is a root, we can factor the polynomial by dividing it by \( n - 1 \). Using synthetic division: ``` 1 | 1 -4 -1 4 | 1 -3 -4 ---------------------- 1 -3 -4 0 ``` This gives us the quotient \( n^2 - 3n - 4 \). Therefore, we can express the original polynomial as: \[ n^3 - 4n^2 - n + 4 = (n - 1)(n^2 - 3n - 4) \] Next, we can factor \( n^2 - 3n - 4 \): To factor, we look for two numbers that multiply to \(-4\) and add to \(-3\). These numbers are \(-4\) and \(1\). Thus, we can factor it as: \[ n^2 - 3n - 4 = (n - 4)(n + 1) \] Now, we can write the complete factorization of the polynomial as: \[ n^3 - 4n^2 - n + 4 = (n - 1)(n - 4)(n + 1) \] Setting each factor equal to zero gives us the solutions: \[ n - 1 = 0 \quad \Rightarrow \quad n = 1 \] \[ n - 4 = 0 \quad \Rightarrow \quad n = 4 \] \[ n + 1 = 0 \quad \Rightarrow \quad n = -1 \] Thus, the complete set of solutions is: \[ n = 1, \quad n = 4, \quad n = -1 \] All solutions are real, and there are no imaginary solutions to this equation.