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July 29, 2014

July 29, 2014

Posted by **kate** on Wednesday, February 4, 2009 at 6:50am.

for what values is # underfined.

- algebra help -
**Reiny**, Wednesday, February 4, 2009 at 8:24amI recall responding to a very similar, perhaps the same question yesterday.

Unless you clarify with brackets where the division starts and ends, I cannot be precise in my answer

the way it stands, only the x^2 is divided, so the expression is undefined for x=0

I am certain you meant:

(x^2+6x+5)/(x^2+8x+15) which then

= ((x+1)(x+5))/((x+3)(x+5))

= (x+1)/(x+3)

so the expression is undefined for x = -3, or x=-5

Some would argue that the expression is undefined for x = -3, and indeterminate for x = -5.

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