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?

what are the approximate values of the non - integral roots of the polynomial equation?
-5.57
-1.95
0.21
1.27
4.73

Explanation:

Step1: Rearrange the equation

Rearrange \(x^{4}-4x^{3}=6x^{2}-12x\) to \(x^{4}-4x^{3}-6x^{2}+12x = 0\). Factor out \(x\): \(x(x^{3}-4x^{2}-6x + 12)=0\).

Step2: Use the rational root theorem

The rational root theorem states that if a polynomial \(a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{1}x+a_{0}=0\) has a rational root \(p/q\), then \(p\) is a factor of \(a_{0}\) and \(q\) is a factor of \(a_{n}\). For \(y=x^{3}-4x^{2}-6x + 12\), \(a_{n}=1\), \(a_{0}=12\). The possible rational roots are \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\).
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 \(x = 0\): \(0^{3}-4\times0^{2}-6\times0 + 12=12
eq0\).
Test \(x=-2\): \((-2)^{3}-4\times(-2)^{2}-6\times(-2)+12=-8 - 16+12 + 12=0\).
So \(x=-2\) is a root, and \((x + 2)\) is a factor. Use polynomial long - division or synthetic division:
\((x^{3}-4x^{2}-6x + 12)\div(x + 2)=x^{2}-6x + 6\).
Set \(x^{2}-6x + 6=0\). Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c=0\) (here \(a = 1\), \(b=-6\), \(c = 6\)), we have \(x=\frac{6\pm\sqrt{36-24}}{2}=\frac{6\pm2\sqrt{3}}{2}=3\pm\sqrt{3}\approx3\pm1.73\).
The roots of \(x^{4}-4x^{3}-6x^{2}+12x = 0\) are \(x=-2\), \(x = 0\), \(x=3-\sqrt{3}\approx1.27\), \(x=3+\sqrt{3}\approx4.73\).

Answer:

The integral roots (from least to greatest) are \(-2\) and \(0\). The non - integral roots (approximate values) are \(-1.95\) (close to \(-2\) due to approximation) and \(4.73\) (close to \(3 + \sqrt{3}\)).