Question

what are the roots of the polynomial equation $x^3 - 7x = 6x - 12$? use a graphing calculator and a system of equations. -4, 1, 3; -3, -1, 4; -6, 6; 1, 3

Explanation:

Step1: Rewrite the equation as a system

We can rewrite the polynomial equation \(x^{3}-7x = 6x - 12\) as a system of two equations:
\(y=x^{3}-7x\) and \(y = 6x-12\). The roots of the original equation are the \(x\)-coordinates of the intersection points of the graphs of these two functions.

Step2: Test the options

We can test each option by substituting the values into the original equation \(x^{3}-7x-(6x - 12)=x^{3}-13x + 12\) and check if the result is \(0\).

• For option \(- 4,1,3\):
• When \(x=-4\): \((-4)^{3}-13\times(-4)+12=-64 + 52+12=0\)
• When \(x = 1\): \(1^{3}-13\times1 + 12=1-13 + 12=0\)
• When \(x = 3\): \(3^{3}-13\times3+12=27-39 + 12=0\)

• For option \(-3,-1,4\):
• When \(x=-3\): \((-3)^{3}-13\times(-3)+12=-27 + 39+12=24

eq0\)

• For option \(-6,6\):
• When \(x=-6\): \((-6)^{3}-13\times(-6)+12=-216+78 + 12=-126

eq0\)

• For option \(1,3\):
• We already saw that \(x=-4\) is also a root, so this option is incomplete.

Answer:

\(-4,1,3\)