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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?

Thanks for your help.

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  • $\begingroup$ what is the problem? both the given expression and the expression you mention after writing "I tried to simply multiply the two rational expressions and found this:", are exactly identical. $\endgroup$ Dec 1, 2022 at 17:58
  • $\begingroup$ Thank you for your reply. The problem that I cannot figure out is why don't the answers from my first attempt match those from the second attempt? Am I doing the math wrong when I cancel out x from the numerator of the first fraction with that in the denominator in the second, and subsequently canceling out (x+4) from the denominator of the first fraction with that in the numerator in the second? $\endgroup$
    – Falcon9
    Dec 1, 2022 at 18:16
  • $\begingroup$ you can only cancel an expression f(x) from the numerator and denominator for values of x such that f(x) is not 0. Otherwise the given original expression is not defined at all, so you cannot perform any operations on it. $\endgroup$ Dec 1, 2022 at 18:47

1 Answer 1

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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 $x = -4, 0, 4$.

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  • $\begingroup$ I believe the author is asking why the expression 4/(x-4) is equivalent but does not reveal the excluded values. $\endgroup$
    – ybakos
    Jan 11, 2023 at 22:12

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