Solve the equation exactly in the complex number system.
-$$x ^ { 3 } - 6 x ^ { 2 } + x - 6 = 0$$
A) $x = \pm 1 , x = 6$
B) $x = 6 , x = \pm i$
C) $x = \pm 6 , x = 1$
D) $x = - 6 , x = \pm i$

Question

Quizplus · Accepted Answer

To solve the equation $x^3 - 6x^2 + x - 6 = 0$ exactly in the complex number system, we can try to factor the equation or use the Rational Root Theorem to find possible rational roots. By inspection or trial and error, we can see that $x = 6$ is a root because substituting $x = 6$ into the equation gives $0$. After finding $x = 6$ as a root, we can perform polynomial division or use synthetic division to divide the original polynomial by $(x - 6)$, which gives us a quadratic equation. The division results in $x^2 + 1 = 0$, which can be rewritten as $x^2 = -1$. Solving this quadratic equation for $x$ gives us the roots $x = \pm i$, where $i$ is the imaginary unit $\sqrt{-1}$.Therefore, the solutions to the equation are $x = 6$ and $x = \pm i$, which matches choice B.

Solve the Equation Exactly in the Complex Number System x3−6x2+x−6=0x ^ { 3 } - 6 x ^ { 2 } + x - 6 = 0x3−6x2+x−6=0

Solve the Equation Exactly in the Complex Number System $x ^ { 3 } - 6 x ^ { 2 } + x - 6 = 0$