Solve the equation below by factoring. $$c ^ { 3 } + 2 c ^ { 2 } - 4 c - 8 = 0$$
A) $c = 0$ and $c = 2$
B) $c = - 2$ , $c = 0$ , and $c = 2$
C) $c = - 2$
D) $c = - 2$ and $c = 2$
E) $c = - 2$ and $c = 0$

Question

Quizplus · Accepted Answer

We can factor the equation as follows: $$c^3 + 2c^2 - 4c - 8 = (c^3 - 2c) + (2c^2 - 4) = c(c^2 - 2) + 2(c^2 - 2) = (c+2)(c-2)(c^2+2c+4) = 0.$$ Therefore, the solutions are $c=-2$ and $c=2.$

Solve the Equation Below by Factoring c3+2c2−4c−8=0c ^ { 3 } + 2 c ^ { 2 } - 4 c - 8 = 0c3+2c2−4c−8=0

Solve the Equation Below by Factoring $c ^ { 3 } + 2 c ^ { 2 } - 4 c - 8 = 0$