Sketch (mentally) y = (x − 2)² and y = (x − 2)³ near x = 2. The first touches the axis at 2 and bounces back up. The second flattens, then crosses. Same zero, different personalities.
The difference is one number: the exponent on the factor. That exponent — the multiplicity — is one of the highest-yield facts on the entire AP Precalc exam. It shows up in MC graph-matching, in FRQ 4 justifications, and in reverse ("here's a graph, write a possible formula").
For a polynomial p, the following say the same thing:
a is a zero of p (that is, p(a) = 0)(x − a) is a linear factor of p(x)p has an x-intercept at (a, 0)A degree-n polynomial has exactly n complex zeros counted with multiplicity, and therefore at most n real zeros.
If p(x) = (x − a)^m · q(x) with q(a) ≠ 0, then a is a zero of multiplicity m.
a (sign of p changes). For m ≥ 3 it flattens as it crosses.a and turns around (sign of p does not change); (a, 0) is a local extremum.[GRAPH: p(x) = (x + 2)(x − 1)²(x − 3) sketched. Crosses at x = −2 (multiplicity 1); touches and bounces at x = 1 (multiplicity 2), where (1, 0) is a local maximum below-to-below... labeled "touch, even multiplicity"; crosses at x = 3 (multiplicity 1). End behavior: degree 4, positive leading coefficient — both ends up. y-intercept at (0, p(0)) = (0, (2)(1)(−3)) = (0, −6).]
Degrees still matter: multiplicities across all factors add up to the degree. In the sketch above: 1 + 2 + 1 = 4. ✓
Some quadratics — like x² + 4 — never touch the x-axis. Their zeros are non-real: x = ±2i. Two facts govern these:
a + bi is a zero, then a − bi is also a zero. Non-real zeros arrive two at a time.Consequences the exam loves:
Two symmetry types, tested algebraically:
f(−x) = f(x) for all x. Graph symmetric about the y-axis. A polynomial is even exactly when all its terms have even exponents (constants count: x⁰).f(−x) = −f(x) for all x. Graph has rotational symmetry about the origin. All terms have odd exponents.Most functions are neither. f(x) = x³ + x is odd; g(x) = x⁴ − 3x²+ 1 is even; h(x) = x² + x is neither (h(−1) = 0 ≠ h(1) = 2 and ≠ −h(1)).
Handy link to zeros: for an odd function, if f(a) = 0 then f(−a) = −f(a) = 0 — zeros come in ± pairs, and if 0 is in the domain, f(0) = −f(0) forces f(0) = 0: every odd function defined at 0 passes through the origin.
Given a graph: read each x-intercept and its cross/touch behavior → write the factors with multiplicities (use the smallest multiplicity consistent with the behavior: cross = 1, cross-with-flattening = 3, touch = 2) → fix the leading constant k with one more point (often the y-intercept):
p(x) = k(x − a)^(m₁)(x − b)^(m₂)…
Problem: p(x) = −2(x + 5)(x − 1)²(x − 4)³. List the zeros, their multiplicities, the degree, and the behavior at each intercept.
Solution: - x = −5, multiplicity 1 → crosses - x = 1, multiplicity 2 → touches (bounces) - x = 4, multiplicity 3 → crosses with flattening - Degree = 1 + 2 + 3 = 6; leading coefficient −2 → even degree, negative leader → both ends down.
Interpretation: Everything about the sketch comes off the factored form in seconds — no plotting.
Problem: The graph of a polynomial p of least possible degree crosses the x-axis at x = −3, touches it at x = 2, and has y-intercept (0, −9). Find a formula for p.
Strategy: Cross → smallest odd multiplicity (1); touch → smallest even multiplicity (2). Then pin the constant k with the y-intercept.
Solution: p(x) = k(x + 3)(x − 2)², degree 1 + 2 = 3.
p(0) = k(3)(−2)² = 12k = −9 → k = −3/4
p(x) = −(3/4)(x + 3)(x − 2)²
Quick check on end behavior: degree 3, negative leading coefficient → up on the left, down on the right; crossing up through −3, touching at 2 from below — consistent with a negative y-intercept between them? p(0) = −9 < 0 sits between the crossing and the touch. The touch at (2, 0) is then a local maximum reaching up to the axis. ✓
Interpretation: A parity fact worth knowing: if all of a polynomial's zeros are real, the number of axis crossings (odd multiplicities) matches the degree's parity — odd + even = odd here, so a graph with exactly one crossing and one touch can only come from an odd-degree polynomial. "Least possible degree" questions and "which degree is possible?" questions (Practice 6) both run on this arithmetic.
Problem: A degree-6 polynomial q with real coefficients has zeros 3 (multiplicity 2) and 1 − 2i. What are the possible numbers of distinct real zeros of q?
Solution: Known so far: 3 (double) and the conjugate pair 1 ± 2i → 2 + 2 = 4 zeros counted with multiplicity; 2 remain. Options for the remaining two: (a) another conjugate pair → distinct real zeros: just {3}, count 1; (b) two real zeros (distinct from each other and from 3, or not): - two new distinct reals → {3, r, s}: count 3 - one new real double (or two equal) → {3, r}: count 2 - both equal to 3 → still {3}: count 1 - one new real... they must come in a pair only if non-real; real zeros can be single? Two remaining slots, each real: any combination works since real zeros need no partner. One slot real and one non-real is impossible (the non-real one would need its conjugate).
Possible counts of distinct real zeros: 1, 2, or 3.
Interpretation: Slots + conjugate-pairing = a counting system. Track "remaining degree" like a budget.
Problem: Determine algebraically whether f(x) = 2x³ − 5x is even, odd, or neither, and state the graphical symmetry.
Solution:
f(−x) = 2(−x)³ − 5(−x) = −2x³ + 5x = −(2x³ − 5x) = −f(x)
f(−x) = −f(x) for all x → odd → rotational symmetry about the origin (180° rotation maps the graph to itself).
Interpretation: Write out f(−x) fully and compare — don't eyeball. The AP wants the algebraic test, and FRQ justifications must show it.
(x−2)² is a shifted parabola — touches. (x−2)³ is a shifted cubic — crosses. Rebuild the rule from these two anytime.x⁴ + x has even degree but is not an even function (the x term breaks symmetry). Fix: even/odd function is about all exponents' parity (algebraic test), not the leading term.(x+3)(x−2)² and −5(x+3)(x−2)² have identical zeros. A graph point (usually the y-intercept) is needed to pin k. Fix: last step of every "write a possible formula" answer — solve for k.p(x) = 4(x − 1)³(x + 2)². At x = −2 the graph of p1. (B). (x + 2)² has even multiplicity → touch-and-turn. (A) needs odd multiplicity; (C) is rational-function behavior; (D) describes the odd-multiplicity-≥3 flattening cross.
2. (C). Zeros: 5, −1, 2 + 3i, and forced conjugate 2 − 3i → at least 4. (A) forgets the conjugate; (B)/(D) overcount.
3. (D). All exponents even (7 = 7x⁰). (A) is odd; (B) mixes parities; (C): |−x| + (−x) = |x| − x ≠ |x| + x.
4. (A). Odd degree → opposite end behaviors → at least one crossing. Non-real zeros pair up, leaving an odd number ≥ 1 of real zeros. (B) impossible for odd degree; (C)/(D) not forced.
5. (C). Cross (min 1) + flattening cross (min 3) + touch (min 2) = 6. (D) forgets flattening needs 3; (B) counts 1+1+2; (A) counts intercepts only.
6. (A). Two touches: even + even = even total degree, and degree ≥ 4. Only (A) is even and ≥ 4. (B)/(C)/(D) are odd — impossible when all multiplicities are even and zeros are real.
g is an odd function with g(4) = −6, and 0 is in the domain of g. Then g(−4) + g(0) =7. (C). g odd → g(−4) = −g(4) = 6; g(0) = 0 (odd + defined at 0). Sum = 6. (A) sign error; (B) forgets to flip; (D) it is determined.
8. (C). Real zeros counted with multiplicity = degree − (number of non-real zeros), and non-real zeros come in pairs → 4, 2, or 0 real zeros with multiplicity. Exactly 3 is impossible. (A) is fine (e.g., one double + two... e.g. (x−1)²(x−2)² → 2 distinct); (B) fine (x⁴ + 1); (D) fine.
9. (A). p(x) = k(x + 1)(x − 2)²; p(0) = k(1)(−2)² = 4k = 8 → k = 2, and the leading coefficient of the expanded cubic is k. (B) sign error; (C) reports the y-intercept; (D) forgets to solve for k.
p(x) = (x² + 9)(x − 2), the real zeros of p are10. (B). x² + 9 = 0 has no real solutions (x = ±3i). Only real zero: x = 2. (A) treats x²+9 like x²−9; (C) lists non-real zeros as "real"; (D) sign error.
h(x) = x⁵ − 4x³ + x. Which is true?11. (C). All exponents odd (5, 3, 1) → h is odd → h(−x) = −h(x) → h(−2) = −h(2). (B) pairs "odd" with the wrong symmetry (y-axis symmetry is even); (A)/(D) misclassify.
12. (FRQ-style, 6 points)
(i) [2 pts] Degree 4. Zeros: −2 (multiplicity 2, graph touches and turns), 1 (multiplicity 1, crosses), 4 (multiplicity 1, crosses).
(ii) [2 pts] Even degree with k < 0 → both ends down: lim x→−∞ f(x) = −∞ and lim x→∞ f(x) = −∞.
(iii) [2 pts] f(2) = k(4)²(1)(−2) = k(16)(1)(−2) = −32k = 32 → k = −1 ✓ (consistent with k < 0). y-intercept: f(0) = −1(2)²(−1)(−4) = −1·4·4 = −16; point (0, −16).
12. (FRQ-style) 🚫 The polynomial f(x) = k(x + 2)²(x − 1)(x − 4) for some constant k < 0.
(i) State the degree, the zeros with multiplicities, and the behavior of the graph at each x-intercept.
(ii) Describe the end behavior using limit notation.
(iii) The graph passes through (2, 32). Find k, then find the y-intercept.
Let p be a polynomial with real coefficients. The table gives selected values of p, and it is known that x = 2 is a zero of p of multiplicity 2, and p has exactly three distinct real zeros: −3, 2, and 5.
| x | −4 | −3 | 0 | 2 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| p(x) | −18 | 0 | 12 | 0 | 3 | 0 | −14 |
(a) (i) Explain why p(x) = 0 has no solution on the interval (2, 5) other than the endpoints. (ii) Is the sign of p positive or negative on (2, 5)? Use a value from the table.
(b) (i) What does the multiplicity of the zero at x = 2 imply about the graph of p at (2, 0)? (ii) Is (2, 0) a local maximum, a local minimum, or neither? Justify using the sign of p near x = 2.
(c) Assuming p has no zeros other than the three given, use the table to determine whether the degree of p is even or odd, and whether its leading coefficient is positive or negative. Justify with the signs of p(−4) and p(6).
(a) [2 pts] (i) [1 pt] p has exactly three distinct real zeros: −3, 2, and 5. Any solution of p(x) = 0 in (2, 5) would be a fourth distinct real zero, contradicting the given information. (ii) [1 pt] p(4) = 3 > 0, and p cannot change sign on (2, 5) without a zero there, so p is positive on all of (2, 5).
(b) [2 pts] (i) [1 pt] Multiplicity 2 is even, so the graph touches the x-axis at (2, 0) and turns around without crossing — p does not change sign at x = 2. (ii) [1 pt] p(0) = 12 > 0 (left of 2) and p(4) = 3 > 0 (right of 2), so p is positive on both sides of x = 2 while p(2) = 0: the point (2, 0) is a local minimum.
(c) [2 pts]
[1 pt] Multiplicities: −3 and 5 are crossings (p changes sign there: p(−4) = −18 < 0 to p(0) = 12 > 0 across −3; p(4) = 3 > 0 to p(6) = −14 < 0 across 5), so they have odd multiplicity — with the double zero at 2, the total degree is odd + even + odd = even.
[1 pt] Since p(−4) < 0 and p(6) < 0, both tails are negative: lim x→−∞ p(x) = −∞ and lim x→∞ p(x) = −∞ (no zeros beyond −3 and 5 means no sign change in the tails). Even degree with both ends down → negative leading coefficient.
1. (B). (x + 2)² has even multiplicity → touch-and-turn. (A) needs odd multiplicity; (C) is rational-function behavior; (D) describes the odd-multiplicity-≥3 flattening cross.
2. (C). Zeros: 5, −1, 2 + 3i, and forced conjugate 2 − 3i → at least 4. (A) forgets the conjugate; (B)/(D) overcount.
3. (D). All exponents even (7 = 7x⁰). (A) is odd; (B) mixes parities; (C): |−x| + (−x) = |x| − x ≠ |x| + x.
4. (A). Odd degree → opposite end behaviors → at least one crossing. Non-real zeros pair up, leaving an odd number ≥ 1 of real zeros. (B) impossible for odd degree; (C)/(D) not forced.
5. (C). Cross (min 1) + flattening cross (min 3) + touch (min 2) = 6. (D) forgets flattening needs 3; (B) counts 1+1+2; (A) counts intercepts only.
6. (A). Two touches: even + even = even total degree, and degree ≥ 4. Only (A) is even and ≥ 4. (B)/(C)/(D) are odd — impossible when all multiplicities are even and zeros are real.
7. (C). g odd → g(−4) = −g(4) = 6; g(0) = 0 (odd + defined at 0). Sum = 6. (A) sign error; (B) forgets to flip; (D) it is determined.
8. (C). Real zeros counted with multiplicity = degree − (number of non-real zeros), and non-real zeros come in pairs → 4, 2, or 0 real zeros with multiplicity. Exactly 3 is impossible. (A) is fine (e.g., one double + two... e.g. (x−1)²(x−2)² → 2 distinct); (B) fine (x⁴ + 1); (D) fine.
9. (A). p(x) = k(x + 1)(x − 2)²; p(0) = k(1)(−2)² = 4k = 8 → k = 2, and the leading coefficient of the expanded cubic is k. (B) sign error; (C) reports the y-intercept; (D) forgets to solve for k.
10. (B). x² + 9 = 0 has no real solutions (x = ±3i). Only real zero: x = 2. (A) treats x²+9 like x²−9; (C) lists non-real zeros as "real"; (D) sign error.
11. (C). All exponents odd (5, 3, 1) → h is odd → h(−x) = −h(x) → h(−2) = −h(2). (B) pairs "odd" with the wrong symmetry (y-axis symmetry is even); (A)/(D) misclassify.
12. (FRQ-style, 6 points)
(i) [2 pts] Degree 4. Zeros: −2 (multiplicity 2, graph touches and turns), 1 (multiplicity 1, crosses), 4 (multiplicity 1, crosses).
(ii) [2 pts] Even degree with k < 0 → both ends down: lim x→−∞ f(x) = −∞ and lim x→∞ f(x) = −∞.
(iii) [2 pts] f(2) = k(4)²(1)(−2) = k(16)(1)(−2) = −32k = 32 → k = −1 ✓ (consistent with k < 0). y-intercept: f(0) = −1(2)²(−1)(−4) = −1·4·4 = −16; point (0, −16).
🎯 Exam tip: When a graph-matching MC gives you four factored formulas, don't expand anything. Check three things in order: (1) end behavior (degree parity + leading sign), (2) cross vs. touch at each intercept (multiplicity parity), (3) y-intercept sign. One of the four survives; it usually takes under a minute.