Q. \( \dfrac{x^2-16}{x-2} \div \dfrac{x^2+3x-4}{x-8} \)

Step-by-step solution

1. Write the original expression.\[
   \dfrac{x^{2}-16}{x-2}\div\dfrac{x^{2}+3x-4}{x-8}
   \]
2. Factor each quadratic where possible. Explain how: use difference of squares for x^{2}-16 and factor the trinomial x^{2}+3x-4 by finding two numbers whose product is -4 and sum is 3 (those numbers are 4 and -1).\[
   x^{2}-16=(x-4)(x+4)
   \]\[
   x^{2}+3x-4=(x+4)(x-1)
   \]

   Replace the quadratics in the expression by their factorizations:

   \[
   \dfrac{(x-4)(x+4)}{x-2}\div\dfrac{(x+4)(x-1)}{x-8}
   \]

3. Record domain restrictions (values that are not allowed). These come from any original denominator and from the divisor being zero. Specifically:
   • x-2 is a denominator in the first fraction, so x cannot equal 2.
   • x-8 is a denominator in the second fraction, so x cannot equal 8.
   • The second fraction as a whole is the divisor; if its numerator x^{2}+3x-4 equals 0 then the divisor is 0 and division would be undefined. So x^{2}+3x-4\neq 0, which means x\neq -4 and x\neq 1.

   Combine these: x cannot be 2, 8, -4, or 1. Keep these exclusions in mind even if factors cancel later.

4. Convert the division to multiplication by the reciprocal. For fractions A/B ÷ C/D we compute (A/B)\times(D/C). Apply that here:\[
   \dfrac{(x-4)(x+4)}{x-2}\times\dfrac{x-8}{(x+4)(x-1)}
   \]
5. Cancel any common nonzero factors between numerator and denominator. There is a common factor (x+4) in a numerator and a denominator. Because x\neq -4 (from the domain restrictions), (x+4) is nonzero and may be canceled.After canceling (x+4) we get:\[
   \dfrac{(x-4)}{x-2}\times\dfrac{x-8}{(x-1)}
   \]
6. Multiply the remaining factors. Multiply numerators together and denominators together:\[
   \dfrac{(x-4)(x-8)}{(x-2)(x-1)}
   \]This is the simplified form, written in factored form. You may expand if desired, but the factored form is usually preferred for clarity about zeros and poles.
7. State the final simplified expression and the domain exclusions.Simplified expression:\[
   \dfrac{(x-4)(x-8)}{(x-2)(x-1)}
   \]

   Domain restrictions (values excluded from the original expression):

   \[
   x\neq -4,\quad x\neq 1,\quad x\neq 2,\quad x\neq 8
   \]

   These exclusions must be kept even though the factor (x+4) canceled algebraically.