Solve the polynomial equation by factoring.
-$$x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12 = 0$$
A) $- 3 , - 4,2$
B) $- 2 , - 3,0$
C) $3,4 , - 2$
D) $2,3 , - 2$

Question

Quizplus · Accepted Answer

To solve the polynomial equation $x^3 - 3x^2 - 4x + 12 = 0$ by factoring, we look for patterns or use techniques such as grouping. Factoring by grouping, we can rearrange the equation as $(x^3 - 3x^2) - (4x - 12) = 0$, which simplifies to $x^2(x - 3) - 4(x - 3) = 0$. Factoring out the common factor $(x - 3)$, we get $(x - 3)(x^2 - 4) = 0$. The term $x^2 - 4$ can be further factored into $(x - 2)(x + 2)$ since it is a difference of squares. Therefore, the equation becomes $(x - 3)(x - 2)(x + 2) = 0$, which has solutions $x = 3$, $x = 2$, and $x = -2$.

Solve the Polynomial Equation by Factoring x3−3x2−4x+12=0x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12 = 0x3−3x2−4x+12=0

Solve the Polynomial Equation by Factoring $x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12 = 0$