https://file.aiquickdraw.com/mathbot/file/1725582040957dpshh19j.jpg

The given rational function in the image is:

\[
r(x) = \frac{x^4 - 12x^3}{x^3 - x^2 - 6x}
\]

We are asked to find the values of \(x\) for which \(r(x) = 0\).

To solve for \(x\), we must set the numerator equal to 0 since a rational function is equal to 0 when the numerator is 0 (provided the denominator is not also 0).

The numerator is:
\[
x^4 - 12x^3 = 0
\]
Factoring the numerator:
\[
x^3(x - 12) = 0
\]
Setting this equal to 0:
\[
x^3(x - 12) = 0
\]
So, the solutions are \(x = 0\) or \(x = 12\).

Next, we need to check that these values do not make the denominator 0. The denominator is:
\[
x^3 - x^2 - 6x
\]
Factoring the denominator:
\[
x(x^2 - x - 6)
\]
Factoring further:
\[
x(x - 3)(x + 2)
\]

Setting the denominator equal to 0 to find restrictions:
\[
x(x - 3)(x + 2) = 0
\]
So, \(x = 0\), \(x = 3\), or \(x = -2\).

Thus, the rational function \(r(x) = 0\) for \(x = 12\), since \(x = 0\) makes the denominator 0, which is undefined. Therefore, \(x = 12\) is the only valid solution.

### Final Answer:
\( r(x) = 0 \) when \( x = 12 \).

Would you like a detailed step-by-step breakdown of any part of this solution?

### Related questions:
1. How would you find vertical asymptotes of the function \(r(x)\)?
2. What are the horizontal asymptotes of the given rational function?
3. How do you factor cubic or higher degree polynomials?
4. Can you describe the behavior of \(r(x)\) as \(x\) approaches infinity?
5. How would you approach solving \(r(x) = 1\)?

### Tip:
Always check for undefined values in rational functions by examining where the denominator is 0 to avoid incorrect solutions.

For what values of x does r(x) = 0, where the rational function is r(x) = (x^4 - 12x^3) / (x^3 - x^2 - 6x)?

Learn how to solve the rational function r(x) = (x^4 - 12x^3) / (x^3 - x^2 - 6x) to find where r(x) = 0. This problem covers key algebraic concepts, including factoring and polynomial equations, with a step-by-step breakdown. Suitable for students in Grades 10-12.

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