Question

use a graphing calculator and a system of equations to find the roots of the equation.
$x^{4}-4x^{3}=6x^{2}-12x$
from least to greatest, what are the integral roots of the equation?
and

Explanation:

Step1: Rearrange the equation

$$x^{4}-4x^{3}-6x^{2}+12x = 0$$
Factor out an \(x\): \(x(x^{3}-4x^{2}-6x + 12)=0\). So \(x = 0\) is a root.

Step2: Use the rational root theorem

For the cubic \(y=x^{3}-4x^{2}-6x + 12\), the possible rational roots are factors of \(12\) (i.e., \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\)).
Test \(x = 2\): \(2^{3}-4\times2^{2}-6\times2 + 12=8-16 - 12 + 12=-8
eq0\).
Test \(x = 3\): \(3^{3}-4\times3^{2}-6\times3 + 12=27-36-18 + 12=-15
eq0\).
Test \(x = 4\): \(4^{3}-4\times4^{2}-6\times4+12=64 - 64-24 + 12=-12
eq0\).
Test \(x=-2\): \((-2)^{3}-4\times(-2)^{2}-6\times(-2)+12=-8 - 16 + 12 + 12=0\).
So \(x=-2\) is a root. Then we can factor the cubic as \((x + 2)(x^{2}-6x + 6)\).
The roots of \(x^{2}-6x + 6\) are \(x=\frac{6\pm\sqrt{36-24}}{2}=3\pm\sqrt{3}\) (not - integral).

Answer:

\(-2\) and \(0\)