A set of real numbers used in place of a variable (x, y, z, etc…) is called the domain. Real numbers are used in place of variables to assist in the functionality of the equations. This excludes any real number that causes the denominator to be zero. The reason a denominator cannot be zero is because when we divide the denominator with the numerator, you cannot divide by zero. You can’t have a value that is undefined.

The two problems that I was assigned are as follows:

1-x2x 2b-22b2-8

In my first expression, I start by transforming it to a rational expression (1-x) (1+x). Since each set contains terms that are perfect squares, I will factor using the difference of squares formula: a²-b²= (a-b) (a+b). In my problem, a = x and gives me: (-x+1) (x+1). Next, I reorder the polynomial from left to right starting with highest order term which is x = 0 As stated above, the domain of an expression is all real numbers except when the expression is undefined. Type equation here.

(1-x)(1+x)x

I simply the FOIL expression by multiplying and combining like terms giving me:

1-x2x

I divide each term in the numerator and denominator. Then I reduce the term by removing the factor of x giving me the expression:

1x-x

The Domain (D) is the set of all Real Numbers, written in set notation that would look like this:
D = {x| x ∈ ℜ} The final answer is: x≠0 Undefined written in set notation. D={x|x ∈ R, x ≠ 0}

In my second expression, I will start by factoring out the GCF (2) from the numerator and -2 and finally 2 from 2b – 2. This leaves me with:

2 (b-1)2b2-8

Next, I factor the denominator using the GCF (2) from 2b2, -8 and 2b2 – 8. This leaves us with the following expression:

2 (b-1)2 (b+2)(b-2)

Each set contains terms that are perfect squares. I reduce the expression by canceling out all common factors from the numerator and denominator.

b-1(b+2)(b-2)

D = {All Real Numbers} Except for my excluded values of 2 and -2.

There were no excluded values in my first expression because x = 0. My second expression contains the excluded values of 2 and -2. If these were used in place of the variables, the denominator would be zero and the expression would become undefined.