# Strange Excluded Values Problem

I am in Algebra II, and my question regarding excluded values came from solving one of my homework questions. It read:

What are the excluded values of the product of: $$\frac{4x}{x^2-16}\times \frac{x+4}{x}$$

I was taught to cancel out factors, multiply, and then find all the zeros in the denominator so that is what I did, coming to the result:

$$\frac{4}{x-4}$$

And so, the only value of x that would result in zero is 4. However, when I submitted my answer, it said I was wrong. So, I reviewed my work and tried solving it another way. I tried to simply multiply the two rational expressions and found this:

$$\frac{(4x)(x+4)}{(x^2-16)(x)}$$

With this new rational expression, the zeros were clear: 0,-4,4. Completely unexpected values.

My question is why is it that different ways of solving the same problem give different answers, and which ones are right?

Since division by zero is undefined, $$\frac{4x}{x^2 - 16} \cdot \frac{x + 4}{x} = \frac{4x(x + 4)}{x(x + 4)(x - 4)}$$ is undefined when $$x = -4, 0, 4$$. The equation $$\frac{4x}{x^2 - 16} \cdot \frac{x + 4}{x} = \frac{4x(x + 4)}{x(x + 4)(x - 4)} = \frac{4}{x - 4}$$ is only valid at those values where the original expression is defined. Thus, the excluded values are $$x = -4, 0, 4$$.
• I believe the author is asking why the expression 4/(x-4) is equivalent but does not reveal the excluded values. Jan 11, 2023 at 22:12