Rational expressions and domains

[PhyiscsisNeat]

Homework Statement

Okay, I have two examples that are confusing me. I am not sure where all the numbers that must be excluded from the denominators so that we're not dividing by zero are coming from.

a) x² + 6x +5 / x² - 25
b) x-7 / x-1 multiplied by x²-1 / 3x-21

Homework Equations

None

The Attempt at a Solution

In a) I factor everything and get (x+5)(x+1) / (x+5)(x-5) and I am left with x+1 / x-5. I know that x cannot be 5, because 5-5=0 and division by zero is undefined. The text is saying that x cannot be -5 as well and I am confused. After factoring and before eliminating common factors, I have an x+5 where if x = -5, I would have 0...is that where the -5 is coming from? Do I consider all x values in the binomials before removing common factors?

In b) after factoring I am left with x-7 / x-1 multiplied by (x+1)(x-1) / 3(x-7). I canceled the x-7 terms and x-1 terms and I am left with x+1 / 3. The text is telling me that x cannot be 1, 7, which is confusing because I eliminated the common factors already.

I guess as I type this out it is making sense...I suppose at this point now that I have typed all this out I would like confirmation if someone could be so kind. It seems like all the x values to be excluded are taken from the binomials in the denominators before I eliminate common factors?

 

[DoItForYourself]

Yes, this is the way it works.

You can understand why, if you calculate the result of (x²+6x+5)/(x²-25) if you put x=-5.

 

[Ray Vickson]

Homework Statement

Okay, I have two examples that are confusing me. I am not sure where all the numbers that must be excluded from the denominators so that we're not dividing by zero are coming from.

a) x² + 6x +5 / x² - 25
b) x-7 / x-1 multiplied by x²-1 / 3x-21

Homework Equations

None

The Attempt at a Solution

In a) I factor everything and get (x+5)(x+1) / (x+5)(x-5) and I am left with x+1 / x-5. I know that x cannot be 5, because 5-5=0 and division by zero is undefined. The text is saying that x cannot be -5 as well and I am confused. After factoring and before eliminating common factors, I have an x+5 where if x = -5, I would have 0...is that where the -5 is coming from? Do I consider all x values in the binomials before removing common factors?

In b) after factoring I am left with x-7 / x-1 multiplied by (x+1)(x-1) / 3(x-7). I canceled the x-7 terms and x-1 terms and I am left with x+1 / 3. The text is telling me that x cannot be 1, 7, which is confusing because I eliminated the common factors already.

I guess as I type this out it is making sense...I suppose at this point now that I have typed all this out I would like confirmation if someone could be so kind. It seems like all the x values to be excluded are taken from the binomials in the denominators before I eliminate common factors?

In both (a) [$x^2 + 6x + \frac{5}{x^2} - 25$] and (b) [$x = \frac{7}{x} -1$] only the point $x=0$ is excluded. Or, maybe, you did not write what you actually meant, in which case you should re-write the expressions to say what you mean. The expression "a + b/c + d" means $a + \frac{b}{c} + d$ when parsed according to official math rules. If you mean $\frac{a+b}{c+d}$ then you need to either use LaTeX (as I have just done) or else use parentheses, like this" "(a+b)/(c+d)".

 

[Mark44]

In a) I factor everything and get (x+5)(x+1) / (x+5)(x-5) and I am left with x+1 / x-5.

In b) after factoring I am left with x-7 / x-1 multiplied by (x+1)(x-1) / 3(x-7).

As Ray mentioned, you need more parentheses. In the first quote above, your first expression isn't correct, even with your use of parentheses. It would be interpreted as $(x + 5) \frac{x+1}{x + 5} (x - 5)$, which is surely not what you meant. A better way to write it on one line would be [(x+5)(x+1)] / [(x+5)(x-5)], so now it's clear which factors are in the numerator and which are in the denominator. As alread mentioned, expressions like x + 1 / x -5 aren't the same as (x + 1)/(x - 5).
Same comments on the second quote.

 

[PhyiscsisNeat]

Hey, thanks for the replies. Yes, everything was in parenthesis but I didn't write it like that in the post. I will do so in the future.

 

[scottdave]

Another thing to take away from this: you cannot "cancel out" expressions which equal zero. You cannot have zero divided by zero