what are the roots of the polynomial equation $x^{3}-6x = 3x^{2}-8$? use a graphing calculator and a system…

Explanation:

Step1: Rearrange the equation

Rewrite $x^{3}-6x = 3x^{2}-8$ as $x^{3}-3x^{2}-6x + 8=0$.

Step2: Test the given - value options

Let's test the values in each option by substituting them into the equation $x^{3}-3x^{2}-6x + 8$. For option 1:

• When $x=-40$: $(-40)^{3}-3\times(-40)^{2}-6\times(-40)+8=-64000 - 4800+240 + 8=-68552\neq0$.
• When $x = - 4$: $(-4)^{3}-3\times(-4)^{2}-6\times(-4)+8=-64-48 + 24+8=-80\neq0$.
• When $x = 5$: $5^{3}-3\times5^{2}-6\times5 + 8=125-75-30 + 8=28\neq0$. For option 2:
• When $x=-5$: $(-5)^{3}-3\times(-5)^{2}-6\times(-5)+8=-125-75 + 30+8=-162\neq0$.
• When $x = 4$: $4^{3}-3\times4^{2}-6\times4 + 8=64-48-24 + 8=0$.
• When $x = 40$: $40^{3}-3\times40^{2}-6\times40+8=64000-4800 - 240+8=58968\neq0$. For option 3:
• When $x=-4$: $(-4)^{3}-3\times(-4)^{2}-6\times(-4)+8=-64-48 + 24+8=-80\neq0$.
• When $x=-1$: $(-1)^{3}-3\times(-1)^{2}-6\times(-1)+8=-1-3 + 6+8=10\neq0$.
• When $x = 2$: $2^{3}-3\times2^{2}-6\times2 + 8=8-12-12 + 8=-8\neq0$. For option 4:
• When $x=-2$: $(-2)^{3}-3\times(-2)^{2}-6\times(-2)+8=-8-12 + 12+8=0$.
• When $x = 1$: $1^{3}-3\times1^{2}-6\times1 + 8=1-3-6 + 8=0$.
• When $x = 4$: $4^{3}-3\times4^{2}-6\times4 + 8=64-48-24 + 8=0$.

Answer:

D. -2, 1, 4