what's the question asking?
the zeroes of the function?
the domain?
in this case, if you want to find the domain
you have to factor the bottom
(5-x)/[(x-3)(x-5)]
and because you know the denominator cannot be zero because that will make the problem undefined, the domain would be
(negative infinite, 3) U (3,5) U (5, infinite)
5-x/x^2-8x+15.
can you show steps on how to solve this? thank you.
2 answers
Does the question say, "simplify" ??
then ...
(5-x)/(x^2-8x+15)
= (5-x)/[(x-3)(x-5)]
= - 1/(x-3) , x ≠ 5, 3
then ...
(5-x)/(x^2-8x+15)
= (5-x)/[(x-3)(x-5)]
= - 1/(x-3) , x ≠ 5, 3
2 answers