I am trying to find the domain for x(x+2)(x+2)/(x-2)(x+2). I canceled the (x+2)'s. The only problem is that my book says the domain cannot equal -2. But I say it cannot equal 2 either (along with -2). (-2-2)(-2+2) is equal to (2-2)(2+2), right? Or what did I do wrong? Thanks!
(-2-2)=0 (-2+2)= -4
(2-2)=0 (2+2)= 4 so they are not the same because the first one is a negative 4 and the second is a positive 4
x(x+2)(x+2)/(x-2)(x+2) line 1
= x(x+2)/(x-2) line 2
you are dealing with explicit vs implicit restrictions.
Some authors of textbooks will list as restrictions all values which make any of the denominators zero.
Other authors, and yours is one of them, will restrict only those factors which were divided out
let me explain:
let x=2
line 1 = 32/0 which is undefined
line 2 = 8/0 which is undefined
so technically line 1 is still equal to line 2 when x=2 and no need to restrict it as (x cannot be equal to 2)
let x=-2
line 1 = 0/0 which is indeterminate (wait till you take Calculus for that one)
line 2 = 0/-4 = 0 a real number
so clearly line 1 is no longer equal to line 2, making x = -2 a "necessary" restriction.
It is a fine distinction. I used to accept both types of answers.
To find the domain of the given expression, we need to consider the values of "x" for which the expression is defined. In this case, we have:
Expression: x(x+2)(x+2)/(x-2)(x+2)
To simplify the expression, we cancel out the common factor of (x+2) in the numerator and denominator:
(x(x+2)(x+2))/(x-2)(x+2) = x(x+2)/(x-2)
Now, we need to determine the values of "x" that make the expression undefined. A rational expression, like the one we have here, is undefined when the denominator equals zero. Therefore, we set the denominator equal to zero and solve for "x":
(x-2) = 0
Solving this equation, we find that x = 2.
So, the expression is undefined when x = 2. Therefore, the domain of the given expression is all the real numbers except x = 2.
Regarding your question about canceling out the (x+2)'s, it's important to note that canceling out terms can only be done when both the numerator and denominator have a common factor that can be eliminated. In this case, canceling out (x+2) was valid since it appeared both in the numerator and denominator. However, when canceling out (x+2), we need to be careful not to cancel out the domain restrictions associated with it. That's why x cannot equal -2 or 2 since these values make the expression undefined.