AP Precalculus: Polynomial Functions
To solve a polynomial inequality: (1) move everything to one side so it reads f(x) > 0, ≥ 0, < 0, or ≤ 0, (2) factor and find the zeros, (3) place the zeros on a number line and test one point in each interval, and (4) keep the intervals that match. Use open circles for < and > and closed circles for ≤ and ≥. A zero of even multiplicity does not change the sign, but it still makes the function equal to 0.
1. What Makes a Function a Polynomial
A polynomial function has the form p(x) = anxn + an−1xn−1 + … + a1x + a0, where n is a whole number and the exponents are all whole numbers. The largest exponent is the degree, the term with that exponent is the leading term, and its coefficient is the leading coefficient. Write terms in descending order, and treat a missing power as having coefficient 0. Roots, variables in denominators, and variables in exponents all rule a function out.
Worked Example: Write p(x) = 5x² − 7x³ + 2 in standard form and name its degree and leading coefficient.
Descending order: p(x) = −7x³ + 5x² + 0x + 2.
Degree 3, leading coefficient −7, constant term 2.
1. Which of the following is a polynomial function?
Explanation: A polynomial allows only whole-number exponents on x. Choice A has x−1, B has x1/2, and D has x in an exponent. Choice C has exponents 5 and 1, with a fraction as a coefficient, which is fine.
2. What are the degree and leading coefficient of p(x) = 4x − 9x5 + x²?
Explanation: The largest exponent is 5, so the degree is 5. The term with that exponent is −9x5, so the leading coefficient is −9, even though it is not written first.
3. What are the degree and leading coefficient of p(x) = (2x − 1)(x + 3)(x² + 1)?
Explanation: The degrees of the factors are 1, 1, and 2, which add to 4. The leading coefficient is the product of the leading coefficients: 2 × 1 × 1 = 2.
2. Relative Extrema, Absolute Extrema, and Inflection
A relative (local) extremum is a peak or valley where the function switches between increasing and decreasing. A polynomial of degree n has at most n − 1 relative extrema. On a closed interval [a, b], the Extreme Value Theorem guarantees both an absolute maximum and an absolute minimum, and each occurs at a relative extremum or at an endpoint. A point of inflection is where concavity changes. A polynomial of degree n ≥ 2 has at most n − 2 points of inflection, and an odd-degree polynomial has at least one. When average rates of change over equal-width steps switch from positive to negative (or the reverse), a relative extremum lies between those steps.
Worked Example: A polynomial has degree 6. How many relative extrema and points of inflection are possible?
Relative extrema: at most 6 − 1 = 5. Inflection points: at most 6 − 2 = 4.
At most 5 relative extrema and at most 4 points of inflection.
1. What is the maximum number of relative extrema a degree 4 polynomial can have?
Explanation: A polynomial of degree n has at most n − 1 relative extrema. For n = 4, that is 4 − 1 = 3.
2. A polynomial function f is defined on the closed interval [−1, 3]. Which statement must be true?
Explanation: By the Extreme Value Theorem, a polynomial on a closed interval always has both an absolute maximum and an absolute minimum. If f is increasing, these occur at the endpoints.
3. Which statement must be true for every polynomial of degree 5?
Explanation: An odd-degree polynomial has at least one point of inflection. The other choices are only upper limits or possibilities. For example, y = x5 has no relative extrema at all.
4. A polynomial’s average rates of change over [0, 1], [1, 2], [2, 3], [3, 4] are 4, 2, −1, −4. What does this suggest?
Explanation: The rates are positive, so the function is increasing, then negative, so it is decreasing. Increasing then decreasing means a relative maximum, and the switch happens between x = 2 and x = 3.
3. Real and Complex Zeros
A polynomial of degree n has exactly n complex zeros, counting repeats, but only the real ones show up as x-intercepts. To find them, set p(x) = 0 and factor completely, by grouping, by a difference of squares, or with the quadratic formula. For a real number a, (x − a) is a factor exactly when p(a) = 0 (the Factor Theorem). If p has real coefficients, nonreal zeros come in conjugate pairs: if a + bi is a zero, so is a − bi. If p has rational coefficients, irrational zeros of the form a + √b also come with their conjugate a − √b.
Worked Example: Find all zeros of f(x) = x4 − 16.
Factor: (x² − 4)(x² + 4) = (x − 2)(x + 2)(x² + 4).
From x² + 4 = 0, x² = −4, so x = ±2i.
Zeros: 2, −2, 2i, −2i (two real, two imaginary).
1. What are the zeros of f(x) = x³ − 2x² − 9x + 18?
Explanation: Group: x²(x − 2) − 9(x − 2) = (x − 2)(x² − 9) = (x − 2)(x − 3)(x + 3). The zeros are 2, 3, and −3.
2. A polynomial with real coefficients has 3 − 2i as a zero. Which number must also be a zero?
Explanation: Nonreal zeros of a real-coefficient polynomial come in conjugate pairs. The conjugate of 3 − 2i is 3 + 2i.
3. A polynomial with rational coefficients has the zeros 2 + √3 and 1 − i. What is the least possible degree?
Explanation: The conjugate 2 − √3 must also be a zero, and so must 1 + i. That is four zeros, so the degree is at least 4.
4. If (x − 2) is a factor of p(x) = x³ + kx² − x − 10, what is k?
Explanation: By the Factor Theorem, p(2) = 0: 8 + 4k − 2 − 10 = 4k − 4 = 0, so k = 1.
4. Multiplicity, Intercept Behavior, and Sign Changes
When a zero appears as a factor more than once, the number of times is its multiplicity, and the multiplicities add up to the degree. At a zero of odd multiplicity the graph crosses the x-axis, and at a zero of even multiplicity it is tangent (touches and turns back). Because polynomials are continuous, the Intermediate Value Theorem says that if f(a) and f(b) have opposite signs, then f has at least one zero between a and b. If the signs match, the number of zeros between them, counted with multiplicity, is even (possibly 0).
Worked Example: Describe the behavior of f(x) = (x + 1)²(x − 3)³ at its zeros.
The degree is 2 + 3 = 5. The zero −1 has multiplicity 2 (even) and the zero 3 has multiplicity 3 (odd).
Tangent to the x-axis at x = −1, and crosses the x-axis at x = 3.
1. For g(x) = x(x − 2)²(x + 4), which statement is correct?
Explanation: The zeros 0 and −4 each have multiplicity 1 (odd), so the graph crosses. The zero 2 has multiplicity 2 (even), so the graph is tangent there.
2. Which polynomial has a zero at 1 of multiplicity 2, a zero at −3 of multiplicity 1, leading coefficient 2, and degree 3?
Explanation: A zero at 1 gives the factor (x − 1), repeated twice, and a zero at −3 gives (x + 3). The leading coefficient 2 multiplies it: 2(x − 1)²(x + 3). Its degree is 3.
3. A polynomial f has f(1) = −4 and f(3) = 5. What must be true?
Explanation: The outputs have opposite signs, and f is continuous, so by the Intermediate Value Theorem there is at least one zero between 1 and 3. There could be more than one.
4. A degree 4 polynomial has p(−1) = 3 and p(2) = 7. What can be concluded about its real zeros between −1 and 2?
Explanation: The signs of p(−1) and p(2) match, so the graph crosses the x-axis an even number of times between them (counting multiplicity), which could be 0, 2, or 4. No more can be concluded.
5. Even and Odd Functions
A function is even if f(−x) = f(x) for every x, and its graph is symmetric about the y-axis. A function is odd if f(−x) = −f(x), and its graph is symmetric about the origin. To test, replace x with −x and simplify. If the result is the original, it is even. If it is the exact opposite, it is odd. If it is neither, the function is neither.
Worked Example: Classify f(x) = x4 − 3x².
f(−x) = (−x)4 − 3(−x)² = x4 − 3x².
f(−x) = f(x), so f is even.
1. Classify g(x) = x³ + 5x.
Explanation: g(−x) = (−x)³ + 5(−x) = −x³ − 5x = −(x³ + 5x) = −g(x), so g is odd.
2. Classify h(x) = x² + x.
Explanation: h(−x) = x² − x. That is not h(x) = x² + x, and it is not −h(x) = −x² − x. So h is neither.
3. An odd function f has f(2) = −6 and f(−3) = 4. What is f(−2) + f(3)?
Explanation: For an odd function, f(−2) = −f(2) = 6 and f(3) = −f(−3) = −4. The sum is 6 + (−4) = 2.
6. End Behavior and Limits
End behavior describes what the outputs do as x moves far to the left or right. Only the leading term matters. Written with limits, limx→∞ p(x) is the value p(x) heads toward as x grows without bound. The rules: even degree means both ends go the same way (up if the leading coefficient is positive, down if negative), and odd degree means the ends go opposite ways (down on the left and up on the right if positive, and the reverse if negative).
Worked Example: State the end behavior of p(x) = −2x4 + x − 1.
The degree 4 is even and the leading coefficient −2 is negative, so both ends fall.
limx→−∞ p(x) = −∞ and limx→∞ p(x) = −∞
1. What is the end behavior of f(x) = x³ − 5x²?
Explanation: The degree 3 is odd and the leading coefficient 1 is positive, so the graph falls on the left and rises on the right: limx→−∞ f(x) = −∞ and limx→∞ f(x) = ∞.
2. What is the end behavior of g(x) = −3x5 + 2x²?
Explanation: The degree 5 is odd and the leading coefficient −3 is negative. The graph rises on the left and falls on the right.
3. The graph of a polynomial falls on both the far left and the far right. Which could be its equation?
Explanation: Falling on both ends means an even degree with a negative leading coefficient. Only y = −x6 + 2x fits. The odd-degree choices have opposite ends, and x4 − 3 rises on both ends.
7. Graphing and Writing Polynomial Functions
Use a seven-step plan to sketch a polynomial: (1) find the degree, (2) count the possible relative extrema and inflection points, (3) find the end behavior, (4) find the zeros by factoring, (5) find each zero’s multiplicity to decide cross or tangent, (6) find the y-intercept with p(0), and (7) draw a smooth continuous curve through the information. To write a function from a graph, run the plan backward: turn each zero into a factor, raise the factor to its multiplicity, and choose the leading coefficient’s sign from the end behavior.
Worked Example: Plan the graph of f(x) = x(x − 2)(x + 3).
Degree 3: at most 2 relative extrema and 1 inflection point. Leading coefficient 1 and odd degree: falls left, rises right. Zeros 0, 2, −3, each multiplicity 1, so it crosses at each. y-intercept: f(0) = 0.
Start low on the left, cross at −3, 0, and 2, and rise on the right.
1. For f(x) = −(x + 1)²(x − 4), which description is correct?
Explanation: The zero −1 has even multiplicity (tangent) and 4 has odd multiplicity (crosses). f(0) = −(1)(−4) = 4. The degree 3 is odd with a negative leading coefficient, so the graph rises on the left and falls on the right.
2. A graph crosses at x = −2, is tangent at x = 1, crosses at x = 3, and rises on both ends. Which function could it be?
Explanation: The multiplicities are odd, even, odd, so the degree is at least 1 + 2 + 1 = 4. Even degree with both ends rising needs a positive leading coefficient, which gives (x + 2)(x − 1)²(x − 3). Choice D is tangent at 3, and B falls on both ends.
3. What is the y-intercept of f(x) = 2(x − 1)(x + 2)(x − 3)?
Explanation: f(0) = 2(−1)(2)(−3) = 2 × 6 = 12, so the y-intercept is (0, 12).
8. The Binomial Theorem
To expand (a + b)n, use the coefficients from row n of Pascal’s triangle (each entry is the sum of the two above it). Rows: 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; 1 5 10 10 5 1. The powers of a go down from n to 0 while the powers of b go up from 0 to n. Treat a minus sign or a coefficient as part of b, and raise it entirely to each power.
Worked Example: Expand (x + 2)4.
Row 4 coefficients: 1, 4, 6, 4, 1. Powers of 2: 1, 2, 4, 8, 16.
x4 + 4(2)x³ + 6(4)x² + 4(8)x + 16 = x4 + 8x³ + 24x² + 32x + 16
1. What is the coefficient of x³y² in the expansion of (x + y)5?
Explanation: Row 5 of Pascal’s triangle is 1, 5, 10, 10, 5, 1. The term with x³y² is the third entry, 10.
2. Which is the expansion of (x − 2y)³?
Explanation: Use row 3 (1, 3, 3, 1) with b = −2y: x³ + 3x²(−2y) + 3x(4y²) + (−8y³) = x³ − 6x²y + 12xy² − 8y³. The signs alternate because b is negative.
3. What is the coefficient of x² in the expansion of (x + 3)4?
Explanation: The x² term uses the third entry of row 4, which is 6, times 3² = 9. So the coefficient is 6 × 9 = 54.
9. Polynomial Inequalities
To solve a polynomial inequality: (1) move everything to one side so it reads f(x) > 0, ≥ 0, < 0, or ≤ 0, (2) factor and find the zeros, (3) place the zeros on a number line and test one point in each interval, and (4) keep the intervals that match. Use open circles for < and > and closed circles for ≤ and ≥. A zero of even multiplicity does not change the sign, but it still makes the function equal to 0.
Worked Example: Solve (x + 1)(x − 2)(x − 5) > 0.
Zeros: −1, 2, 5. Test x = −2: (−)(−)(−) is negative. Test x = 0: (+)(−)(−) is positive. Test x = 3: (+)(+)(−) is negative. Test x = 6: positive.
Solution: (−1, 2) ∪ (5, ∞)
1. Solve x² − 4x − 5 ≤ 0.
Explanation: Factor: (x − 5)(x + 1) ≤ 0. The parabola opens upward, so it is at or below zero between its zeros. Including the endpoints gives [−1, 5].
2. Solve x³ − 9x < 0.
Explanation: Factor: x(x − 3)(x + 3) < 0. Zeros: −3, 0, 3. For x > 3 the product is positive, and the sign alternates: negative on (0, 3), positive on (−3, 0), negative on (−∞, −3). The negative intervals are (−∞, −3) and (0, 3).
3. Solve (x − 2)²(x + 4) ≥ 0.
Explanation: The factor (x − 2)² is never negative and equals 0 at x = 2, so the sign depends on (x + 4). It is nonnegative when x ≥ −4, and x = 2 is allowed because the inequality includes 0. The solution is [−4, ∞).
Common Mistakes to Avoid
- Counting x-intercepts instead of zeros. A degree n polynomial has n complex zeros, but repeated and nonreal zeros do not appear as separate x-intercepts.
- Forgetting the conjugate. When a real-coefficient polynomial has a zero like 2 + 3i, its conjugate 2 − 3i must also be a zero, so the degree is at least 2 from that pair alone.
- Mixing up tangent and cross. Odd multiplicity crosses the x-axis, and even multiplicity touches it and turns around.
- Reading end behavior from the wrong term. Only the highest-degree term controls end behavior, even if it is not written first.
- Treating “at most” as “exactly.” A degree n polynomial has at most n − 1 relative extrema and at most n − 2 inflection points, not always that many.
- Dropping the sign inside the binomial theorem. In (x − 2y)³, the whole quantity −2y is raised to each power, so the signs alternate.
- Ignoring even-multiplicity zeros in inequalities. They do not change the sign, but they do make the function 0, which matters for ≤ and ≥.
Frequently Asked Questions
How is a zero different from an x-intercept?
A zero is a number x that makes p(x) = 0, and it can be real or complex. An x-intercept is a point (a, 0) on the graph, so only real zeros give x-intercepts.
How can I tell the degree from a graph?
Count the relative extrema and inflection points. A graph with k relative extrema has degree at least k + 1, and the end behavior tells you whether the degree is even or odd.
Why does a polynomial with real coefficients have conjugate pairs?
When you substitute a + bi and a − bi into a polynomial with real coefficients, the imaginary parts cancel in pairs, so if one of them makes the polynomial 0 the other must as well.
Where can I practice more problems like these?
The Polynomial Functions quiz in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.