Solve the polynomial equation in the complex numbers.
-$$x ^ { 3 } + 8 x ^ { 2 } + 16 x + 8 = 0$$
A) $\{ 1 , - 1 , - 8 \}$
B) $\{ 2 , - 6 + \sqrt { 5 } , - 6 - \sqrt { 5 } \}$
C) $\{ - 2 , - 3 + \sqrt { 5 } , - 3 - \sqrt { 5 } \}$
D) $\{ - 2 , - 6 + \sqrt { 8 } , - 6 - \sqrt { 8 } \}$

Question

Quizplus · Accepted Answer

The given polynomial can be factored as a sum of cubes: $x^3 + 8x^2 + 16x + 8 = (x + 2)^3 = 0$, which gives a root of $x = -2$. The quadratic factor can be solved to find the other two roots, which are $-3 + \sqrt{5}$ and $-3 - \sqrt{5}$.

Solve the Polynomial Equation in the Complex Numbers x3+8x2+16x+8=0x ^ { 3 } + 8 x ^ { 2 } + 16 x + 8 = 0x3+8x2+16x+8=0

Solve the Polynomial Equation in the Complex Numbers $x ^ { 3 } + 8 x ^ { 2 } + 16 x + 8 = 0$