Question

How do I determine the restricted values in this expression?

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

Answer 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

Answer 2

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?

Answer 3

((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))