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?

Explanation:

Step1: Rearrange the equation

Rewrite $x^{4}-4x^{3}=6x^{2}-12x$ as $x^{4}-4x^{3}-6x^{2}+12x = 0$. Then factor out an $x$: $x(x^{3}-4x^{2}-6x + 12)=0$.

Step2: Group - factor the cubic part

Group the terms in $x^{3}-4x^{2}-6x + 12$: $(x^{3}-4x^{2})-(6x - 12)=x^{2}(x - 4)-6(x - 2)$. Another way is to try to find roots by trial - and - error. We check possible rational roots using the Rational Root Theorem. The possible rational roots of $x^{3}-4x^{2}-6x + 12$ are factors of $12$, i.e., $\pm1,\pm2,\pm3,\pm 4,\pm6,\pm12$.
When $x = 2$: $2^{3}-4\times2^{2}-6\times2 + 12=8-16 - 12 + 12=-8
eq0$.
When $x=-2$: $(-2)^{3}-4\times(-2)^{2}-6\times(-2)+12=-8 - 16 + 12 + 12=0$.
So $(x + 2)$ is a factor of $x^{3}-4x^{2}-6x + 12$. Using polynomial long - division or synthetic division: $x^{3}-4x^{2}-6x + 12=(x + 2)(x^{2}-6x + 6)$.

Step3: Solve the factored equation

The original equation $x(x^{3}-4x^{2}-6x + 12)=0$ becomes $x(x + 2)(x^{2}-6x + 6)=0$.
Set each factor equal to zero:

• $x=0$ gives one root.
• $x + 2=0$ gives $x=-2$.
• For $x^{2}-6x + 6=0$, use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b=-6$, and $c = 6$. Then $x=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times6}}{2\times1}=\frac{6\pm\sqrt{36 - 24}}{2}=\frac{6\pm\sqrt{12}}{2}=\frac{6\pm2\sqrt{3}}{2}=3\pm\sqrt{3}$.

The integral roots from least to greatest are $x=-2$ and $x = 0$.

Answer:

$-2,0$