Question

solving polynomial equations using technology
using a system to solve a polynomial equation with irrational roots
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

First, we rewrite the given equation \(x^{4}-4x^{3}=6x^{2}-12x\) in standard form by moving all terms to one side:
\(x^{4}-4x^{3}-6x^{2}+12x = 0\)

Step2: Factor the polynomial

Factor out the greatest common factor (GCF) from the first two terms and the last two terms:
\(x^{3}(x - 4)-6x(x - 2)=0\)
Wait, maybe a better way is to factor by grouping or factor out \(x\) first. Let's factor out \(x\) first:
\(x(x^{3}-4x^{2}-6x + 12)=0\)
Now, factor the cubic polynomial \(x^{3}-4x^{2}-6x + 12\) by grouping. Group the first two terms and the last two terms:
\(x^{2}(x - 4)-6(x - 2)=0\)
Hmm, maybe we can try rational root theorem. The possible rational roots of \(x^{3}-4x^{2}-6x + 12\) are factors of 12 over factors of 1, so \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\).
Let's test \(x = 0\): Plug into the original equation \(0^{4}-4\times0^{3}-6\times0^{2}+12\times0=0\), so \(x = 0\) is a root.
Test \(x = 2\): \(2^{4}-4\times2^{3}-6\times2^{2}+12\times2=16-32 - 24 + 24=-16 eq0\)
Test \(x = 3\): \(3^{4}-4\times3^{3}-6\times3^{2}+12\times3=81-108 - 54 + 36=-45 eq0\)
Test \(x = 6\): \(6^{4}-4\times6^{3}-6\times6^{2}+12\times6=1296-864 - 216 + 72=288 eq0\)
Test \(x=-2\): \((-2)^{4}-4\times(-2)^{3}-6\times(-2)^{2}+12\times(-2)=16 + 32-24 - 24=0\). Wait, \((-2)^4=16\), \(-4\times(-2)^3=-4\times(-8)=32\), \(-6\times(-2)^2=-6\times4 = - 24\), \(12\times(-2)=-24\). So \(16 + 32-24 - 24=0\). So \(x=-2\) is a root? Wait, no, let's check the original equation \(x^{4}-4x^{3}=6x^{2}-12x\). If \(x = 0\), left side: \(0-0 = 0\), right side: \(0 - 0=0\), so \(x = 0\) is a root. If \(x = 2\), left side: \(16-32=-16\), right side: \(24 - 24 = 0\), so \(-16 eq0\). If \(x = 6\), left side: \(1296-864 = 432\), right side: \(6\times36-12\times6=216 - 72 = 144\), \(432 eq144\). Wait, maybe I made a mistake in factoring. Let's start over.
Original equation: \(x^{4}-4x^{3}-6x^{2}+12x = 0\)
Factor out \(x\): \(x(x^{3}-4x^{2}-6x + 12)=0\)
Now, let's factor the cubic \(x^{3}-4x^{2}-6x + 12\). Let's group as \((x^{3}-4x^{2})+(-6x + 12)=x^{2}(x - 4)-6(x - 2)\). Not helpful. Let's try to use the rational root theorem again. Possible roots: \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\).
Test \(x = 2\): \(8-16-12 + 12=-8 eq0\)
Test \(x = 3\): \(27-36-18 + 12=-15 eq0\)
Test \(x = 6\): \(216-144-36 + 12=48 eq0\)
Test \(x = 1\): \(1-4-6 + 12=3 eq0\)
Test \(x=-1\): \(-1-4 + 6 + 12=13 eq0\)
Test \(x = \sqrt{...}\)? Wait, maybe we can factor the original equation as \(x^{4}-4x^{3}-6x^{2}+12x=0\)
\(x(x^{3}-4x^{2}-6x + 12)=0\)
Wait, maybe there was a typo in the equation. Wait, the original equation is \(x^{4}-4x^{3}=6x^{2}-12x\), let's move all terms to left: \(x^{4}-4x^{3}-6x^{2}+12x = 0\)
Factor by grouping:
\(x^{4}-4x^{3}-6x^{2}+12x=x^{3}(x - 4)-6x(x - 2)\). No, that's not helpful. Wait, maybe the equation is \(x^{4}-4x^{3}=6x^{2}-12x\), let's divide both sides by \(x\) (assuming \(x eq0\)): \(x^{3}-4x^{2}=6x - 12\), then \(x^{3}-4x^{2}-6x + 12=0\)
Now, let's try to factor this cubic. Let's use synthetic division. Let's try \(x = 2\):
Coefficients: 1 | -4 | -6 | 12
Bring down 1. Multiply by 2: 2. Add to -4: -2. Multiply by 2: -4. Add to -6: -10. Multiply by 2: -20. Add to 12: -8. Not zero.
Try \(x = 6\):
1 | -4 | -6 | 12
Bring down 1. Multiply by 6: 6. Add to -4: 2. Multiply by 6: 12. Add to -6: 6. Multiply by 6: 36. Add to 12: 48. Not zero.
Try \(x = \sqrt{...}\)? Wait, maybe the equation is \(x^{4}-4x^{3}=6x^{2}-12x\), let's rearrange as \(x^{4}-4x^{3}-6x^{2}+12x = 0\), then factor as \(x(x^{3}-4x^{2}-6x + 12)=0\). Wait, maybe there is a mistake in the problem statement, or maybe I misread it. Wait, the user wrote \(z^{4}-4z^{3}=6z^{2}-12z\)? Maybe it's \(x\) instead of \(z\). Let's assume the equation is \(x^{4}-4x^{3}-6x^{2}+12x = 0\). Let's factor out \(x\): \(x(x^{3}-4x^{2}-6x + 12)=0\). Now, let's try to factor the cubic. Let's group terms:
\(x^{3}-4x^{2}-6x + 12=(x^{3}-4x^{2})+(-6x + 12)=x^{2}(x - 4)-6(x - 2)\). Not helpful. Wait, maybe the equation is \(x^{4}-4x^{3}=6x^{2}-12x\), let's move all terms to left: \(x^{4}-4x^{3}-6x^{2}+12x = 0\), then factor as \(x(x^{3}-4x^{2}-6x + 12)=0\). Let's try to find integer roots. Let's test \(x = 0\): works. \(x = 2\): \(16-32-24 + 24=-16 eq0\). \(x = 3\): \(81-108-54 + 36=-45 eq0\). \(x = 6\): \(1296-864-216 + 72=288 eq0\). \(x=-2\): \(16 + 32-24-24=0\). Oh! \(x=-2\): \((-2)^4=16\), \(-4\times(-2)^3=-4\times(-8)=32\), \(-6\times(-2)^2=-6\times4=-24\), \(12\times(-2)=-24\). So \(16 + 32-24-24=0\). So \(x=-2\) is a root. Wait, but when we plug \(x=-2\) into the original equation: left side: \((-2)^4-4\times(-2)^3=16 + 32=48\). Right side: \(6\times(-2)^2-12\times(-2)=6\times4 + 24=24 + 24=48\). So \(x=-2\) is a root. And \(x = 0\) is a root. Wait, but let's check \(x = 2\): left side: \(16-32=-16\), right side: \(24-24=0\), not equal. \(x = 6\): left side: \(1296-864=432\), right side: \(6\times36-72=216-72=144\), not equal. Wait, maybe there are two integer roots: \(x = 0\) and \(x=-2\)? Wait, no, when we factored out \(x\), we have \(x = 0\) as a root, and then we found \(x=-2\) is a root of the cubic. Wait, let's do synthetic division with \(x=-2\) on the cubic \(x^{3}-4x^{2}-6x + 12\):
Coefficients: 1 | -4 | -6 | 12
Bring down 1. Multiply by -2: -2. Add to -4: -6. Multiply by -2: 12. Add to -6: 6. Multiply by -2: -12. Add to 12: 0. So the cubic factors as \((x + 2)(x^{2}-6x + 6)\). So the original polynomial factors as \(x(x + 2)(x^{2}-6x + 6)=0\). So the roots are \(x = 0\), \(x=-2\), and the roots of \(x^{2}-6x + 6=0\) (which are irrational: \(x=\frac{6\pm\sqrt{36 - 24}}{2}=\frac{6\pm\sqrt{12}}{2}=3\pm\sqrt{3}\)). So the integer roots are \(x=-2\) and \(x = 0\). Wait, but when we tested \(x = 0\), it worked, and \(x=-2\) worked. Let's confirm \(x = 0\): left side \(0-0=0\), right side \(0 - 0=0\). \(x=-2\): left side \(16 + 32=48\), right side \(24 + 24=48\). So the integer roots are \(-2\) and \(0\).

Answer:

The integer roots from least to greatest are \(-2\) and \(0\).