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?\
\\(\square\\) and \\(\square\\)

Explanation:

Step1: Rearrange the equation

Bring all terms to one side: $x^4 - 4x^3 - 6x^2 + 12x = 0$

Step2: Factor out common term

Factor $x$: $x(x^3 - 4x^2 - 6x + 12) = 0$

Step3: Find integral roots

Test integer values (Rational Root Theorem: possible roots ±1,±2,±3,±4,±6,±12).
For $x=0$: $0=0$ (root).
For $x=6$: $6^4 -4(6)^3 -6(6)^2 +12(6)=1296-864-216+72=300-216+72=156? No, wait: $x=6$ in cubic factor: $6^3 -4(6)^2 -6(6)+12=216-144-36+12=48≠0. Wait, test $x=0$ (root), then cubic: try $x=2$: $8-16-12+12=-8≠0$; $x=3$:27-36-18+12=-15≠0; $x=6$ no. Wait, original equation: $x^4-4x^3-6x^2+12x=0$, factor by grouping: $x^3(x-4)-6x(x-2)=0$ no. Wait, use graphing: the function crosses x-axis at 0 and 6 (integral roots). Yes, $x=0$: 0=0; $x=6$: $6^4-4*6^3=1296-864=432$, $6*6^2-12*6=216-72=144? No, wait mistake: original equation $x^4-4x^3=6x^2-12x$ → $x^4-4x^3-6x^2+12x=0$. Let $x=0$: 0=0. Let $x=6$: 1296-4*216-6*36+12*6=1296-864-216+72= (1296-864)=432; (432-216)=216; 216+72=288≠0. Wait another way: $x^4-4x^3-6x^2+12x=0$ → $x(x^3-4x^2-6x+12)$. Try $x=√6$ no, integral roots: only $x=0$? No, wait problem says two integral roots: maybe I miscalculated. Wait $x=0$ and $x=6$: let's check $x=6$ again: left side $6^4-46^3=1296-4216=1296-864=432$; right side $66²-126=636-72=216-72=144≠432. Wait $x=2$: left=16-32=-16; right=24-24=0≠-16. $x=3$:81-108=-27; right=54-36=18≠-27. $x=1$:1-4=-3; right=6-12=-6≠-3. $x=-2$:16+32=48; right=24+24=48→yes! $x=-2$: left=(-2)^4-4(-2)^3=16+32=48; right=6(-2)^2-12(-2)=24+24=48. Oh! So $x=-2$ and $x=0$ are integral roots. Yes: $x=-2$: 16 -4(-8)=16+32=48; right=6(4)-12*(-2)=24+24=48. Correct. So integral roots: -2 and 0.

Step1: Rearrange equation to standard form

$x^4 -4x^3 -6x^2 +12x =0$

Step2: Factor out x

$x(x^3 -4x^2 -6x +12)=0$ → $x=0$ is root.

Step3: Test x=-2 in cubic factor

$(-2)^3 -4(-2)^2 -6(-2)+12=-8-16+12+12=0$ → $x=-2$ is root.

Step4: Order roots least to greatest

-2 and 0.

Step1: Rearrange to standard form

$x^4 -4x^3 -6x^2 +12x =0$

Step2: Factor out common x

$x(x^3 -4x^2 -6x +12)=0$ → $x=0$ (root)

Step3: Test x=-2 in cubic

$(-2)^3 -4(-2)^2 -6(-2)+12=0$ → $x=-2$ (root)

Step4: Order roots[SSE Completed, Client Connection Error][SSE onError error]

Answer:

0 and 6