AP PRECALCULUS — PREREQUISITE REVIEW
Polynomial Expressions
Notes — Prerequisite Topic 2
💡 Learning Objectives
By the end of this lesson you will be able to:
- Identify the degree, leading coefficient, and constant term of a polynomial
- Add, subtract, and multiply polynomial expressions fluently
- Expand products using distribution, FOIL, and the box/area model
- Factor quadratic trinomials of the form ax² + bx + c
- Recognize and factor special products (difference of squares, perfect square trinomials)
- Apply the quadratic formula to polynomial problems and interpret the discriminant
1. Polynomial Vocabulary
A polynomial is a sum of terms, where each term is a real-number coefficient times a non-negative integer power of the variable. Polynomials are the building blocks of most functions you will study in AP Precalculus.
Term | Meaning | Example |
|---|---|---|
Degree | highest power of the variable | 3x⁵ − x + 7 has degree 5 |
Leading coefficient | coefficient of the highest-degree term | 3 in 3x⁵ − x + 7 |
Constant term | term with no variable | 7 in 3x⁵ − x + 7 |
Monomial | polynomial with one term | −4x³ |
Binomial | polynomial with two terms | x² − 9 |
Trinomial | polynomial with three terms | 2x² + 3x − 5 |
2. Adding, Subtracting, and Multiplying Polynomials
2.1 Addition and subtraction
Combine only like terms — terms that share the same variable raised to the same power.
📘 Example — Adding and subtracting
Simplify (3x³ − 5x + 2) + (x³ + 4x² − 7) − (2x³ − 3x + 1).
= 3x³ − 5x + 2 + x³ + 4x² − 7 − 2x³ + 3x − 1
= (3 + 1 − 2)x³ + 4x² + (−5 + 3)x + (2 − 7 − 1)
= 2x³ + 4x² − 2x − 6.
2.2 Multiplying polynomials
Multiplication uses the distributive property: every term of the first polynomial multiplies every term of the second. The area (box) model is a clean way to organize the work and makes sure no pairs are missed.
Area model for (x + 3)(x + 2). Each inner rectangle represents one of the four partial products; adding them gives x² + 5x + 6.
📘 Example — Multiplying a binomial by a trinomial
Expand (x − 2)(x² + 3x − 4).
Distribute each term of (x − 2) across the trinomial:
x(x² + 3x − 4) = x³ + 3x² − 4x.
−2(x² + 3x − 4) = −2x² − 6x + 8.
Sum: x³ + 3x² − 4x − 2x² − 6x + 8 = x³ + x² − 10x + 8.
💡 Special products worth memorizing
(a + b)² = a² + 2ab + b² — perfect square trinomial
(a − b)² = a² − 2ab + b²
(a + b)(a − b) = a² − b² — difference of squares
Recognizing these patterns on sight speeds up both expanding and factoring.
3. Factoring Quadratic Trinomials
Factoring reverses multiplication. Given ax² + bx + c, you want to write it as a product of two binomials.
3.1 When the leading coefficient is 1
For x² + bx + c, look for two numbers p and q such that p · q = c and p + q = b. Then:
x² + bx + c = (x + p)(x + q).
📘 Example — Factoring x² + 7x + 12
We need two numbers whose product is 12 and whose sum is 7.
The pair 3 and 4 works (3 · 4 = 12 and 3 + 4 = 7).
Therefore x² + 7x + 12 = (x + 3)(x + 4).
3.2 When the leading coefficient is not 1
For ax² + bx + c with a ≠ 1, use the ac-method:
- Step 1: multiply a · c.
- Step 2: find two numbers that multiply to ac and add to b.
- Step 3: split the middle term using those numbers.
- Step 4: factor by grouping.
📘 Example — Factoring 6x² + 7x − 3
Here a · c = 6 · (−3) = −18 and b = 7.
Two numbers that multiply to −18 and add to 7 are 9 and −2.
Split: 6x² + 9x − 2x − 3.
Group: (6x² + 9x) + (−2x − 3) = 3x(2x + 3) − 1(2x + 3).
Factor: (2x + 3)(3x − 1).
4. Special Factoring Patterns
Parabolas grouped by the discriminant Δ = b² − 4ac. Δ controls how many real zeros a quadratic has, which matches how (or whether) its trinomial factors over the real numbers.
Pattern | Factors as | Example |
|---|---|---|
a² − b² | (a + b)(a − b) | x² − 25 = (x + 5)(x − 5) |
a² + 2ab + b² | (a + b)² | x² + 10x + 25 = (x + 5)² |
a² − 2ab + b² | (a − b)² | 4x² − 12x + 9 = (2x − 3)² |
⚠️ Sum of squares does not factor over the reals
x² + 9 cannot be factored into two linear factors with real coefficients. Over the complex numbers, x² + 9 = (x + 3i)(x − 3i) — you will return to this in the Complex Numbers review.
5. The Quadratic Formula Revisited
💡 Using the quadratic formula to factor
When a trinomial resists the ac-method, the quadratic formula always locates its roots:
x = ( −b ± √(b² − 4ac) ) / (2a).
If those roots are r₁ and r₂, the original trinomial factors as a(x − r₁)(x − r₂).
📘 Example — Factoring when inspection fails
Factor 2x² − 7x + 3.
Formula: x = (7 ± √(49 − 24)) / 4 = (7 ± 5) / 4.
Roots: x = 3 and x = ½.
Factored form: 2(x − 3)(x − ½) = (x − 3)(2x − 1).
6. Sign Analysis of Factored Polynomials
Once a polynomial is factored, a sign chart (or sign analysis) tells you where the polynomial is positive and where it is negative — essential for solving polynomial inequalities and for later work on rational functions.
Sign chart for (x + 2)(x − 3). The zeros at x = −2 and x = 3 split the real line into three intervals; the product alternates sign as x passes each zero.
7. Summary
- Add and subtract polynomials by combining like terms only
- Multiply with the distributive property, FOIL, or an area model
- Memorize the three special products: (a ± b)² and (a + b)(a − b)
- Factor x² + bx + c by finding a pair (p, q) with p · q = c, p + q = b
- Factor ax² + bx + c with the ac-method and grouping
- When stuck, get the roots via the quadratic formula and write a(x − r₁)(x − r₂)