Solve the polynomial equation by factoring and then using the zero product principle.
-$$3 x - 4 = 12 x ^ { 3 } - 16 x ^ { 2 }$$
A) $\left\{ - \frac { 1 } { 2 } , \frac { 1 } { 2 } , \frac { 4 } { 3 } \right\}$
B) $\left\{ 0 , \frac { 4 } { 3 } \right\}$
C) $\left\{ - \frac { 1 } { 4 } , \frac { 1 } { 4 } , \frac { 4 } { 3 } \right\}$
D) $\left\{ - \frac { 1 } { 2 } , \frac { 1 } { 2 } , \frac { 3 } { 4 } \right\}$

Question

Quizplus · Accepted Answer

First, rearrange the equation to set it to zero: $0 = 12x^3 - 16x^2 - 3x + 4$. Factoring by grouping, we get $4x^2(3x - 4) - 1(3x - 4) = 0$, which simplifies to $(3x - 4)(4x^2 - 1) = 0$. Further factoring gives $(3x - 4)(2x - 1)(2x + 1) = 0$, leading to the solutions $x = \frac{4}{3}, \frac{1}{2}, -\frac{1}{2}$.

Solve the Polynomial Equation by Factoring and Then Using the Zero