AP Precalculus: Equivalent Representations (Drill 1)

Question 1. Which of the following is equivalent to \( \dfrac{x^2 + 3x - 10}{x - 2} \) for all x ≠ 2?

• A) x − 5
• B) x + 5 ✓
• C) x + 3
• D) \( x^2 + 3x - 5 \)

Explanation: Choice B is correct. Factoring the numerator: x² + 3x − 10 = (x − 2)(x + 5). Dividing by (x − 2) cancels the common factor, leaving x + 5 for all x ≠ 2. Choice A is incorrect because x − 5 would require the numerator to factor as (x − 2)(x − 5) = x² − 7x + 10, which does not match the given numerator. Choice C is incorrect because x + 3 results from incorrectly using the coefficient of x rather than properly factoring; the middle term 3x cannot be isolated as the answer without factoring the full expression. Choice D is incorrect because it leaves the expression in an unsimplified quadratic form and subtracts 2 from the constant erroneously, division by (x − 2) eliminates one linear factor, not one constant.

Question 2. Using polynomial long division, which of the following is equivalent to \( \dfrac{2x^3 - x^2 + 4x - 3}{x - 1} \)?

• A) \( 2x^2 + x + 5 + \dfrac{2}{x - 1} \) ✓
• B) \( 2x^2 - 3x + 1 \)
• C) \( 2x^2 + x + 5 \)
• D) \( 2x^2 + x + 3 + \dfrac{2}{x - 1} \)

Explanation: Choice A is correct. Performing polynomial long division of 2x³ − x² + 4x − 3 by (x − 1): Step 1, 2x³ ÷ x = 2x²; multiply: 2x²(x − 1) = 2x³ − 2x²; subtract to get x² + 4x − 3. Step 2, x² ÷ x = x; multiply: x(x − 1) = x² − x; subtract to get 5x − 3. Step 3, 5x ÷ x = 5; multiply: 5(x − 1) = 5x − 5; subtract to get remainder 2. The result is 2x² + x + 5 + 2/(x − 1). Confirm: substituting x = 1 into the numerator gives 2 − 1 + 4 − 3 = 2, confirming a nonzero remainder of 2. Choice B is incorrect due to sign errors in the middle steps of the division. Choice C is incorrect because it omits the remainder term; since the remainder is 2 (not 0), the division is not exact. Choice D is incorrect because of an arithmetic error in computing the constant term of the quotient, the correct value is 5, not 3.

Question 3. A company's total production cost (in dollars) is modeled by C(x) = 3x + 500, where x is the number of units produced (x > 0). The average cost per unit is \( A(x) = \dfrac{3x + 500}{x} \). Which equivalent form of A(x) best reveals the long-run average cost behavior as production increases without bound?

• A) \( A(x) = \dfrac{3x + 500}{x} \)
• B) \( A(x) = 3 + \dfrac{500}{x} \) ✓
• C) \( A(x) = 3\!\left(1 + \dfrac{500}{3x}\right) \)
• D) \( A(x) = \left(3 + \dfrac{500}{x}\right) \cdot 1 \)

Explanation: Choice B is correct. Dividing each term of the numerator by x gives A(x) = 3 + 500/x. This form immediately reveals that as production x increases without bound, the term 500/x approaches zero, and the average cost per unit approaches $3. The fixed cost of $500 has diminishing per-unit impact at higher production volumes. Choice A is algebraically equivalent but does not make this long-run behavior transparent, the approach to a limiting value of $3 is not visible in the original fraction. Choice C factors out 3 from the expression but buries the key term inside a fraction-within-a-fraction, making the limiting behavior harder to identify. Choice D is a trivial algebraic restatement that adds no interpretive clarity.

Question 4. Which of the following is an equivalent form of \( f(x) = \dfrac{x^2 - 4}{x + 2} \) that reveals a feature not visible in the original expression?

• A) \( f(x) = \dfrac{x^2}{x + 2} - \dfrac{4}{x + 2} \)
• B) \( f(x) = \dfrac{(x - 2)(x + 2)}{x + 2} \)
• C) f(x) = x − 2, x ≠ −2 ✓
• D) f(x) = x + 2 − 4/(x + 2) + 2/(x + 2)

Explanation: Choice C is correct. Factoring the numerator as (x − 2)(x + 2) and canceling (x + 2) from the denominator reveals that f is equivalent to the linear function y = x − 2 everywhere except at x = −2, where there is a hole. This equivalent form reveals that the graph is a straight line with a removable discontinuity, a feature that is completely hidden in the original rational expression. Choice A splits the fraction into two terms but reveals nothing new about the function's behavior or graph. Choice B shows the factored numerator but does not complete the cancellation, so it is still a rational expression with the same hidden features as the original. Choice D contains algebraic errors, the terms do not simplify correctly, and even if they did, the form would not clearly reveal the linear nature of the function.

Question 5. A student rewrites \( g(x) = \dfrac{x^3 + 8}{x + 2} \) using polynomial long division and obtains \( g(x) = x^2 - 2x + 4 \) for all real x. A second student says this rewrite is incorrect because the original function is undefined at x = −2. Which of the following best evaluates the second student's claim?

• A) The second student is wrong: polynomial long division always produces an expression valid for all real numbers.
• B) The second student is wrong: since the remainder is 0, the two expressions are identical everywhere.
• C) The second student is correct: g(x) and \( x^2 - 2x + 4 \) are completely different functions with no algebraic connection.
• D) The second student is correct: the rewritten form \( x^2 - 2x + 4 \) is equivalent to g(x) for all x ≠ −2, but at x = −2 the original is undefined, creating a hole that the polynomial form does not show. ✓

Explanation: Choice D is correct. While x³ + 8 = (x + 2)(x² − 2x + 4) factors exactly with remainder 0, the original rational expression g(x) is undefined at x = −2 because division by zero is undefined. The polynomial x² − 2x + 4 is defined at x = −2, substituting gives (−2)² − 2(−2) + 4 = 4 + 4 + 4 = 12, so the two expressions differ at that point. The correct statement is that g(x) = x² − 2x + 4 for all x ≠ −2, but g has a hole at x = −2 that the polynomial form conceals. The first student's error was claiming the equivalence holds "for all real x." Choice A is incorrect because polynomial long division of rational expressions always requires specifying where the denominator equals zero. Choice B commits the same error, a zero remainder confirms exact division for x ≠ −2, not for all real x. Choice C overstates the second student's position; the two expressions agree everywhere except at x = −2.