Q. ( dfrac{3x^2-18x}{x^2-2x-24} )

Given expression

\( \dfrac{3x^2-18x}{x^2-2x-24} \)

1. Factor the numerator by taking out the greatest common factor.

   Both terms in the numerator, 3x^2 and −18x, share the common factor 3x. Factor it out:

   \( 3x^2-18x = 3x\,(x-6) \)

   Explanation: 3x times x gives 3x^2, and 3x times −6 gives −18x.

2. Factor the denominator as a quadratic trinomial.

   We look for two numbers whose product is −24 and whose sum is −2. These numbers are −6 and +4. So the denominator factors as:

   \( x^2-2x-24 = (x-6)(x+4) \)

   Explanation: (x−6)(x+4) expands to x^2 +4x −6x −24 = x^2 −2x −24.

3. Write the rational expression with the factored numerator and denominator.

   \( \dfrac{3x^2-18x}{x^2-2x-24} = \dfrac{3x\,(x-6)}{(x-6)(x+4)} \)

4. Cancel the common factor where allowed.

   The factor (x−6) appears in both numerator and denominator, so it cancels provided x−6 ≠ 0, i.e. x ≠ 6. After cancellation we obtain:

   \( \dfrac{3x\,(x-6)}{(x-6)(x+4)} = \dfrac{3x}{x+4} \quad\text{for } x \neq 6 \)

   Explanation: Cancellation of common nonzero factors produces an equivalent expression for all x where the factor is not zero.

5. State domain restrictions coming from the original expression.

   The original denominator is zero when x satisfies x^2−2x−24 = 0, i.e. when x = 6 or x = −4. Therefore the original expression is undefined at x = 6 and x = −4. Even though cancellation yields a simpler form, the values x = 6 and x = −4 remain excluded from the domain.

   Notes: At x = 6 the original expression is of the indeterminate form 0/0 (a removable discontinuity or hole). At x = −4 the original expression has a nonzero numerator over zero denominator (a vertical asymptote).

6. Final simplified form with domain restriction.

   \( \dfrac{3x^2-18x}{x^2-2x-24} = \dfrac{3x}{x+4} \quad\text{with } x \neq -4,\,6. \)

   Explanation: The simplified expression represents the same values as the original for every x except where the original is undefined (x = 6 and x = −4).