What is a polynomial function?
A polynomial function is f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where the powers are whole numbers (0, 1, 2, …) and aₙ ≠ 0.
- Degree n: the highest power.
- Leading coefficient aₙ: the number in front of the highest power.
- Constant term a₀ = f(0): the y-intercept.
Names: degree 0 constant, 1 linear, 2 quadratic, 3 cubic, 4 quartic, 5 quintic. Not polynomials: 1/x, √x, 2ˣ.
Finite differences: for equally spaced x, the nth differences of a degree-n polynomial are constant (and equal aₙ × n!). This lets you find the degree from a table.
End behaviour
For very large |x| the leading term aₙxⁿ wins, so it decides where the ends go.
| aₙ > 0 | aₙ < 0 | |
|---|---|---|
| n even | ↖ … ↗ (both up) | ↙ … ↘ (both down) |
| n odd | ↙ … ↗ (left down, right up) | ↖ … ↘ (left up, right down) |
In symbols: if n is odd and aₙ > 0, then as x → −∞, y → −∞ and as x → +∞, y → +∞.
Symmetry: only even powers → even function (y-axis symmetry); only odd powers → odd function (origin symmetry).
Zeros, factors and multiplicity
Factor theorem: f(r) = 0 exactly when (x − r) is a factor. Remainder theorem: dividing f(x) by (x − r) leaves remainder f(r).
If (x − r) appears k times, r has multiplicity k:
- k = 1: the graph crosses straight through.
- k = 2 (even): the graph touches and bounces back.
- k = 3 (odd, > 1): the graph crosses but flattens like y = x³.
Sketching from factored form: (1) plot zeros, (2) y-intercept f(0), (3) end behaviour from the leading term, (4) cross or bounce at each zero.
Writing an equation from a graph: read zeros and multiplicities, write y = a(x − r₁)(x − r₂)…, then use one more point to find a. A family of polynomials y = k(x − r₁)(x − r₂) shares zeros but has different k.
Turning points and solving polynomial equations
A degree-n polynomial has at most n − 1 turning points (local max/min) and at most n real zeros.
Solving f(x) = 0:
- Take out a common factor.
- Try integer factors of a₀ (rational root test: ± factors of a₀ ÷ factors of aₙ).
- When f(r) = 0, divide by (x − r) (long or synthetic division).
- Solve the quadratic left over.
Inequalities f(x) > 0: find zeros, then use the graph or a sign table to see where the curve is above the axis.
Complex zeros
The fundamental theorem of algebra: a degree-n polynomial has exactly n zeros, counting multiplicity and complex zeros.
If the coefficients are real, complex zeros come in conjugate pairs a ± bi. They do not show as x-intercepts. Example: x³ − x² + 4x − 4 = (x − 1)(x² + 4) has zeros 1, 2i, −2i; the graph crosses the axis only once.
Rates of change
Average rate of change from x = a to x = b: [f(b) − f(a)] ÷ (b − a), the slope of the secant.
- Linear: the rate is the same everywhere (the slope).
- Quadratic: the rate changes steadily; first differences grow by a constant.
- Higher degree: the rate changes faster.
Instantaneous rate at x = a: shrink the interval (a, a + h) with h very small; the secant becomes the tangent. Near a turning point it is 0. This leads to derivatives in calculus.
Try it: build a box
Take an A4-size sheet (about 30 cm × 21 cm). Cut equal squares of side x from the corners and fold up a box. Try x = 2, 4, 6 cm and measure the volume. Compare with V = x(30 − 2x)(21 − 2x). Which x gives the biggest box? (About 4 cm: a turning point.)
Key formulas and definitions
- f(x) = aₙxⁿ + … + a₁x + a₀
- Factor theorem: f(r) = 0 ⇔ (x − r) is a factor
- Remainder of f(x) ÷ (x − r) = f(r)
- Real zeros ≤ n; turning points ≤ n − 1
- Average rate = [f(b) − f(a)] / (b − a)
- Instantaneous rate ≈ [f(a + h) − f(a)] / h, h → 0
Worked examples
1. Give the degree, leading coefficient and end behaviour of f(x) = −2x⁴ + 3x − 1.
Degree 4 (even), aₙ = −2 (negative). Both ends go down: ↙ … ↘.
2. Find the zeros of f(x) = x(x − 2)²(x + 3) and say cross or bounce.
x = 0 (×1, cross), x = 2 (×2, bounce), x = −3 (×1, cross).
3. Is (x − 2) a factor of x³ − 3x² + 4?
f(2) = 8 − 12 + 4 = 0. Yes.
4. Solve x³ − 6x² + 11x − 6 = 0.
Try x = 1: 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor. Divide: x² − 5x + 6 = (x − 2)(x − 3). Zeros 1, 2, 3.
5. A cubic has zeros −1, 2, 4 and passes through (0, 16). Find it.
y = a(x + 1)(x − 2)(x − 4). At x = 0: a(1)(−2)(−4) = 8a = 16, a = 2. y = 2(x + 1)(x − 2)(x − 4).
6. Find the average rate of change of f(x) = x³ from x = 1 to x = 3.
(27 − 1) ÷ (3 − 1) = 26 ÷ 2 = 13.
7. Find all zeros of x³ + x² + 9x + 9.
Group: x²(x + 1) + 9(x + 1) = (x + 1)(x² + 9). Zeros −1, 3i, −3i.
8. Solve (x + 2)(x − 1)(x − 3) > 0.
Zeros −2, 1, 3; odd degree, a > 0 so the graph starts below. Above the axis on −2 < x < 1 and x > 3.
Common mistakes
- Reading the ends from the first written term instead of the highest power.
- Thinking every zero is a crossing; even multiplicity only touches.
- Saying a degree-4 polynomial must have 4 x-intercepts. Some zeros may be complex or repeated.
- Sign slip: the factor (x + 2) gives the zero x = −2, not 2.