AP Precalculus: Polynomial Functions and Complex Zeros (Drill 4)

Question 1. Let p(x) = −x(x − 4)(x + 2). On which of the following intervals is p(x) ≥ 0?

• A) (−∞, −2] and [4, ∞)
• B) [−2, 0] and [4, ∞)
• C) (−∞, −2] and [0, 4] ✓
• D) [0, 4] only

Explanation: Choice C is correct. The zeros of p(x) = −x(x − 4)(x + 2) are x = −2, x = 0, and x = 4 (each multiplicity 1). These divide the real line into four intervals. Since the leading term is −x³, the polynomial is negative as x → +∞ and positive as x → −∞. Testing each interval: On (−∞, −2), test x = −3: p(−3) = −(−3)(−7)(−1) = (3)(−7)(−1) = 21 > 0. On (−2, 0), test x = −1: p(−1) = −(−1)(−5)(1) = −5 0. On (4, ∞), test x = 5: p(5) = −(5)(1)(7) = −35 < 0. Therefore p(x) ≥ 0 on (−∞, −2] and [0, 4], including the zeros at the endpoints. Choice A is incorrect because the leading coefficient is negative (the leading term is −x³), so the polynomial is negative, not positive, as x → +∞. Students who ignore the negative sign on the leading factor commonly choose (4, ∞) as a positive interval. Choice B is incorrect because [−2, 0] is a negative interval (test point x = −1 gives p(−1) = −5 < 0). This choice incorrectly combines the correct interval [0, 4] with the negative interval [−2, 0]. Choice D is incorrect because it omits the interval (−∞, −2], which is also positive for this polynomial. Choosing only [0, 4] misses the other positive region on the left side of the number line.

Question 2. Let p(x) = x(x − 3)(x + 1)(x − 5). How many of the intervals in the partition of the real number line formed by the zeros of p contain points where p(x) < 0?

• A) One
• B) Two ✓
• C) Three
• D) Four

Explanation: Choice B is correct. The zeros of p(x) = x(x − 3)(x + 1)(x − 5) are x = −1, x = 0, x = 3, and x = 5 (in order). These create five intervals: (−∞, −1), (−1, 0), (0, 3), (3, 5), (5, ∞). The leading term is x⁴ (positive), so p is positive as x → ±∞, meaning the outermost intervals are positive. Testing each interval: (−∞, −1): test x = −2: (−2)(−5)(−1)(−7) = 70 > 0. (−1, 0): test x = −0.5: (−0.5)(−3.5)(0.5)(−5.5) ≈ −4.8 0. (3, 5): test x = 4: (4)(1)(5)(−1) = −20 0. Exactly two intervals, (−1, 0) and (3, 5), contain points where p(x) < 0. Choice A is incorrect because it finds only one negative interval, missing either (−1, 0) or (3, 5). Choice C is incorrect because it counts three of the five intervals as negative; this would be true if the leading coefficient were negative (as in a −x⁴ polynomial), but the leading term here is positive x⁴. Choice D is incorrect because a degree-4 polynomial with positive leading coefficient must be positive on the outermost intervals (as x → ±∞), so it cannot be negative on four of five intervals.

Question 3. Let p(x) = (x + 3)(x² − 2x − 15). How many distinct real zeros does p have?

• A) 1
• B) 2 ✓
• C) 3
• D) 4

Explanation: Choice B is correct. The key step is to factor x² − 2x − 15 completely. We need two numbers that multiply to −15 and add to −2: those are −5 and +3, giving x² − 2x − 15 = (x − 5)(x + 3). Therefore p(x) = (x + 3)(x − 5)(x + 3) = (x + 3)²(x − 5). The zeros are x = −3 (multiplicity 2) and x = 5 (multiplicity 1). There are exactly 2 distinct real zeros: −3 and 5. Choice A is incorrect because it would mean p crosses (or touches) the x-axis at only one x-value. But p(5) = 0 and p(−3) = 0 are two different x-values, so at least two zeros exist. Choice C is incorrect because it results from failing to factor x² − 2x − 15, then treating (x + 3) and (x² − 2x − 15) as three separate factors contributing three distinct zeros. After factoring, the zero x = −3 appears twice; it is a repeated zero, not a second distinct zero. Choice D is incorrect because a degree-3 polynomial can have at most 3 zeros counting multiplicity, p(x) = (x + 3)²(x − 5) has exactly 3 zeros counting multiplicity (x = −3 twice, x = 5 once), all real. There is no fourth zero.

Question 4. A polynomial function p has real coefficients and degree 4. It is known that 2 + 3i is a zero of p, and that p has exactly two distinct real zeros. Which of the following must be true?

• A) 2 − 3i is also a zero of p, and p has no other non-real zeros. ✓
• B) −2 + 3i is also a zero of p, because complex zeros come in sign-opposite pairs.
• C) p has a total of three non-real zeros, since the real part, imaginary part, and conjugate each contribute a separate zero.
• D) p could have only one real zero, since the remaining three zeros could all be non-real.

Explanation: Choice A is correct. By the Complex Conjugate Pairs Theorem, whenever a polynomial with real coefficients has a non-real zero a + bi (b ≠ 0), its conjugate a − bi must also be a zero. Therefore 2 − 3i is a zero of p. Since p has degree 4, it has exactly 4 zeros counting multiplicity. We now account for all of them: 2 + 3i and 2 − 3i are a conjugate pair (2 non-real zeros), and the problem states p has exactly 2 distinct real zeros (2 real zeros). That accounts for all 4 zeros, leaving no room for additional non-real zeros. Choice B is incorrect because the Complex Conjugate Pairs Theorem states that the complex conjugate flips the sign of the imaginary part, giving 2 − 3i, not −2 + 3i. Flipping the sign of the real part produces a different complex number that is not guaranteed to be a zero. Choice C is incorrect because a complex zero such as 2 + 3i is a single zero, not three separate zeros. The conjugate pair {2 + 3i, 2 − 3i} contributes exactly 2 zeros total, not 3. Choice D is incorrect because the problem explicitly states that p has exactly two distinct real zeros, so it cannot have only one real zero. Non-real zeros in a real-coefficient polynomial must appear in conjugate pairs, so the number of non-real zeros must be even, meaning a degree-4 polynomial can have 0, 2, or 4 non-real zeros.

Question 5. A polynomial function p has degree 3, real coefficients, and leading coefficient 1. The zeros of p are x = 2, x = −1, and x = 4. What is p(0)?

• A) −8
• B) 8 ✓
• C) −2
• D) 5

Explanation: Choice B is correct. Since p has degree 3, leading coefficient 1, and zeros at x = 2, x = −1, and x = 4, the polynomial in factored form is p(x) = (x − 2)(x + 1)(x − 4). Evaluating at x = 0: p(0) = (0 − 2)(0 + 1)(0 − 4) = (−2)(1)(−4). Multiplying: (−2)(1) = −2, then (−2)(−4) = 8. Therefore p(0) = 8. Choice A is incorrect because −8 results from dropping the negative sign on the factor (0 − 4), computing (−2)(1)(4) = −8 instead of (−2)(1)(−4) = 8. This is a sign error: students who forget that (0 − 4) = −4 (not +4) arrive at −8. Choice C is incorrect because −2 results from multiplying only the first two factors and ignoring the third: (0 − 2)(0 + 1) = (−2)(1) = −2. Choice D is incorrect because 5 results from adding the zeros: 2 + (−1) + 4 = 5. The sum of the zeros equals the negative of the coefficient of the x² term (by Vieta's formulas), which is not the same as evaluating the polynomial at x = 0.