SOLUTION: problem is to find all numbers for which each rational expression is undefined. If the rational expression is defined for all real numbers, it says, so state. x+3 over(x-2)(x+5)

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Question 85401This question is from textbook Introductrory & Intermediate Algebra for college students
: problem is to find all numbers for which each rational expression is undefined. If the rational expression is defined for all real numbers, it says, so state.
x+3 over(x-2)(x+5)
7 over X^2 + 81 this one is also in fraction form This question is from textbook Introductrory & Intermediate Algebra for college students

Found 2 solutions by bucky, scianci:
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
Given:
.

.
Note that this function will be undefined if a division by zero is involved. (Division by
zero is not allowed in algebra.) So if either of the factors in the denominator is equal to
zero, the entire denominator is zero and this makes the function undefined. Now we need to
determine the values of x that will make the denominator equal to zero. Note that if
then one of the factors is zero and the entire denominator becomes zero. Solve
this equation and you get . This tells us that x cannot equal +2 or the function
is undefined. Similarly cannot equal zero. So set and when
you solve that you get . Therefore, x cannot equal -5 or the entire denominator
will become 0 because will be zero. In summary, there are two rational numbers
x = +2 and x = -5 for which the function is undefined.
.
Next problem:
.

.
Notice that x^2 is always a positive number or zero. Therefore, the denominator of this
function cannot be zero. (It will be at least 81 and that occurs when x = 0.) You can
also tell that the denominator cannot be zero by setting up the equation:
.

.
Subtract 81 from both sides of this equation and you get:
.

.
But this is impossible because whenever you square a positive or a negative number the
answer is always positive. So there is no way you can multiply a number by itself and
get minus 81 as the answer. From this you can conclude that the denominator of this function
cannot equal zero. So the function is defined for all real numbers assigned to x.
.
Hope this helps to clarify things for you regarding these two problems.

Answer by scianci(186) (Show Source):
You can put this solution on YOUR website!
Look for the zeros of the denominator. Since it's in factored form, it's easy.
x - 2 = 0 and x + 5 = 0
Solving each of these for x will give you the values you need.