posted by kate on .
for what values is # underfined.
I 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
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.