How do I determine the restricted values in this expression?

((x^2+4)/(x^2-3x+2))/((x^4-16)/(x-1))

there are three denominators in the problem

... two fractions , divided by each other

restricted values are usually the result of attempting
... to divide by zero (BIG no-no)

any value that results in a zero denominator is restricted

My choices are

0
1
-4
4
2
-2
-1

So I know 0 is one then so would 2 and -2 also be ones?

((x^2+4)/(x^2-3x+2))/((x^4-16)/(x-1))

= ((x^2+4)/(x^2-3x+2)) * (x-1)/(x^4-16)
= (x^2+4)(x-1) / ((x-1)(x-2)(x^2-4)(x^2+4))
= (x^2+4)(x-1) / ((x-1)(x-2)^2(x+2)(x^2+4))

The original fractions eliminate (x-1) and (x-2)
The final result also disallows (x+2)
for all x not equal to 1, the fraction is
1/((x-2)^2(x+2))

To determine the restricted values in the given expression, you need to analyze the denominator(s) of the expression. Restricted values occur when the denominator(s) of a fraction equal zero, as division by zero is undefined.

Let's break down the expression and identify the denominators:

((x^2+4)/(x^2-3x+2))/((x^4-16)/(x-1))

The denominators in this expression are:
1. (x^2-3x+2) in the first fraction.
2. (x-1) in the second fraction.

To find the restricted values, set each denominator equal to zero and solve for x.

For the first denominator: x^2-3x+2 = 0
You can factor the quadratic equation: (x-2)(x-1) = 0
Now set each factor equal to zero:
x - 2 = 0 or x - 1 = 0
x = 2 or x = 1

Therefore, the restricted values for the first fraction are x = 2 and x = 1.

For the second denominator: x - 1 = 0
Set the equation equal to zero:
x = 1

Therefore, the restricted value for the second fraction is x = 1.

Combining the results, the restricted value for the expression ((x^2+4)/(x^2-3x+2))/((x^4-16)/(x-1)) is x = 1.