How do you simplify #(z^2+x-1)/(z^2-3z+2)# and find the excluded values?

1 Answer
Jan 26, 2017
  1. Factor both numerator and denominator (separately).
  2. Reduce (if possible) by cancelling out common factors.
  3. To find excluded values, set the denominator of the expression (from step 1) equal to zero; solve for #z#.

Explanation:

When simplifying a fraction such as this, we always want to start by factoring the numerator and denominator. Multiplicative factors may divide out, and they certainly help us find the excluded values.

Let's start with the numerator: #z^2+x-1#. Can we find two numbers that add to 1 and multiply to -1? Not easily. They do exist, but they involve radicals. For now, let's leave this as a trinomial.

Next, we look at the denominator: #z^2-3z+2#. This time, we can easily find two numbers that add to -3 and multiply to 2—those numbers are -2 and -1. We can then factor the denominator as follows:

#z^2-3z+2" "=" "(z-2)(z-1)#

So, without factoring the numerator, our expression is now

#(z^2+x-1)/((z-2)(z-1))#

In this form, it is easy to deduce the excluded values, because an excluded value is any value for the variable which makes the expression undefined, such as values that create division by zero. So we look at our denominator and see if there are any #z#-values that make it 0.

Since our denominator is now written as a product of two factors, it will only equal 0 when one of the factors is 0. (In math terms, if the product #a * b = 0,# then either #a=0# or #b=0#.) We then simply set each of the (denominator) factors equal to 0 and solve for #z#.

You can write this in math as follows:

If
#"        "(z-2)(z-1)=0#
then
#"        "z-2=0" ""or"" "z-1=0#
#=>"         "z=2"                    "z=1#

Both #z=1# and #z=2# create division by zero. If #z=1# or #z=2#, our expression #(z^2+x-1)/(z^2-3z+2)# is undefined. Thus, our excluded values are 1 and 2:

#z!=1, 2#.