what are the roots of the polynomial equation $x^{3}-7x = 6x - 12$? use a graphing calculator and a system…

Explanation:

Step1: Rearrange the equation

First, rewrite the equation $x^{3}-7x = 6x - 12$ as $x^{3}-13x + 12=0$.

Step2: Test the given values

Let's test the values in each option by substituting them into the equation $x^{3}-13x + 12$. For $x=-6$: $(-6)^{3}-13\times(-6)+12=-216 + 78+12=-126\neq0$. For $x = 6$: $6^{3}-13\times6 + 12=216-78 + 12=150\neq0$. For $x=-4$: $(-4)^{3}-13\times(-4)+12=-64 + 52+12=0$. For $x = 1$: $1^{3}-13\times1+12=1-13 + 12=0$. For $x = 3$: $3^{3}-13\times3+12=27-39 + 12=0$. For $x=-3$: $(-3)^{3}-13\times(-3)+12=-27+39 + 12=24\neq0$. For $x=-1$: $(-1)^{3}-13\times(-1)+12=-1 + 13+12=24\neq0$.

Answer:

B. -4, 1, 3