Canceling a Common Factor in a Rational Expression
During a peer-tutoring session, two students simplify the rational expression (x^2 − 9)/(x^2 + x − 12), which appears in a cost-per-unit model where x is the number of hundreds of units produced and x ≥ 0. Maya writes (x − 3)/(x + 4) and states that the domain is all real numbers except x = −4. Devon writes (x + 3)/(x + 4) but removes the restriction x ≠ 3 because 'the factor x − 3 canceled.' The group checks x = 0 and gets 3/4 from both the original expression and Devon's version. At x = 3, the original expression is 0/0, while Devon's version gives 6/7. At x = −4, both versions have denominator zero. The tutor asks you to settle the disagreement without evaluating at only one convenient point, and to explain what the cancellation step actually permits in a model restricted to x ≥ 0.
1. Factor the original numerator and denominator. State the correct simplified expression and its domain restrictions, and identify the mistake in each student's work.
Read the worked answer
The numerator factors as x^2 − 9 = (x − 3)(x + 3). The denominator factors as x^2 + x − 12 = (x + 4)(x − 3). The common factor is x − 3, not x + 3. Canceling it is valid only where x − 3 ≠ 0, so the correct simplification is (x + 3)/(x + 4), with x ≠ 3 and x ≠ −4. Under the cost model's condition x ≥ 0, the practical domain is x ≥ 0 with x ≠ 3. Maya canceled x + 3 even though the denominator has no matching factor x + 3; her expression is not equivalent, and at x = 0 it gives −3/4 rather than the original 3/4. Devon has the correct simplified form but wrongly dropped x ≠ 3: the original expression is undefined at x = 3, and canceling removes a hole in the graph, not the domain restriction.
2. Devon uses the simplified expression to compute C(3) = 6/7. Is that value a valid value of the original model? If the model must be continuous at x = 3, what should be specified?
Read the worked answer
No. In the original expression, substituting x = 3 gives 0/0, so C(3) is undefined. The simplified expression agrees with the original everywhere except at x = 3, where it assigns 6/7. The graph therefore has a removable discontinuity, or hole, at (3, 6/7). Assigning C(3) = 6/7 is valid only if the model explicitly defines that value as a continuous extension. Without that added definition, 6/7 is not a value of the original model. The excluded value x = −4 cannot be filled in this way: the original numerator is 7 and the simplified numerator is −1 there, so the graph has a vertical asymptote, not a hole.