Overview
Polynomial operations and factoring form the algebraic backbone of SAT Advanced Math. The SAT tests adding, subtracting, and multiplying polynomials, recognizing special product patterns, applying the Factor Theorem to find roots, and using polynomial long division. Every Digital SAT includes 2–3 questions testing the Factor Theorem and polynomial zeros. Mastery of GCF factoring and pattern recognition dramatically reduces solve time.
Key Points
1. Adding and Subtracting Polynomials
Combine like terms (same variable, same degree):
(3x² + 2x - 1) + (x² - 5x + 4) = 4x² - 3x + 3
(3x² + 2x - 1) - (x² - 5x + 4) = 2x² + 7x - 5
Critical rule: When subtracting, distribute the negative sign to every term in the second polynomial.
2. Multiplying Polynomials
Use the distributive property (FOIL for binomials):
(2x + 3)(x - 4) = 2x² - 8x + 3x - 12 = 2x² - 5x - 12
For three or more terms, multiply two at a time.
3. Special Products (Must Memorize)
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
(a + b)(a - b) = a² - b² ← Difference of Squares
a³ + b³ = (a + b)(a² - ab + b²) ← Sum of Cubes
a³ - b³ = (a - b)(a² + ab + b²) ← Difference of Cubes
4. Factoring Strategy (Step-by-Step)
- Factor out GCF first — always the first step
- Count remaining terms:
- 2 terms: check difference of squares or sum/difference of cubes
- 3 terms: trinomial factoring (trial and error or AC method)
- 4 terms: factoring by grouping
Factoring by grouping (4-term polynomial):
x³ + 2x² + 9x + 18
= (x³ + 2x²) + (9x + 18)
= x²(x + 2) + 9(x + 2)
= (x + 2)(x² + 9)
Binomial GCF (SAT hard version):
3(x - 3)² + 7(x - 3)
= (x - 3)[3(x - 3) + 7]
= (x - 3)(3x - 2)
5. Factor Theorem
If f(k) = 0, then (x - k) is a factor of f(x). Conversely, if (x - k) is a factor, then f(k) = 0.
SAT application: A point (k, 0) on the graph of f(x) means k is a root, so (x - k) is a factor.
If f(2) = 0, then (x - 2) is a factor of f(x).
6. Polynomial Long Division
Used to divide f(x) by (x - k) or other divisors:
Divide x³ - 2x² + x - 3 by (x - 1):
x³ - 2x² + x - 3 = (x - 1)(x² - x + 0) + (-3)
Quotient: x² - x Remainder: -3
Remainder Theorem: The remainder when f(x) is divided by (x - k) equals f(k).
7. End Behavior
For polynomial f(x) = aₙxⁿ + … :
| Leading coefficient | Degree | Left end | Right end |
|---|---|---|---|
| Positive | Even | Up | Up |
| Negative | Even | Down | Down |
| Positive | Odd | Down | Up |
| Negative | Odd | Up | Down |
Pitfalls and Common Mistakes
Pitfall 1: Not distributing the negative sign when subtracting polynomials
WRONG: (3x² + 2x) - (x² + 5x) = 3x² + 2x - x² + 5x
RIGHT: (3x² + 2x) - (x² + 5x) = 3x² + 2x - x² - 5x = 2x² - 3x
Fix: Always distribute the minus sign to every term in the parentheses being subtracted.
Pitfall 2: Incomplete GCF factoring
WRONG: 6x³ + 9x² → 3(2x³ + 3x²) (GCF of 3 only, missed x²)
RIGHT: 6x³ + 9x² → 3x²(2x + 3)
Fix: Check whether a power of the variable is also part of the GCF.
Pitfall 3: Misapplying the Factor Theorem direction Students apply it backwards: because (x + 2) is a factor, they say f(2) = 0 instead of f(–2) = 0. Fix: (x – k) is a factor → f(k) = 0. Note the sign: (x + 2) = (x – (–2)), so k = –2.
Pitfall 4: Forgetting to verify factoring by multiplying back Fix: Distribute your factored answer and confirm it matches the original. Alternatively, substitute x = 1 into both the original and factored form to check.
Pitfall 5: Confusing difference of squares with sum of squares a² + b² does NOT factor over the real numbers. Only a² – b² = (a+b)(a–b). Fix: Sum of squares is prime (irreducible) over real numbers.
Related Entries
- Quadratic_Equations
- Quadratic_Functions_Parabolas
- Rational_Expressions_Equations
- Systems_Nonlinear_Equations
- Functions_Concepts_Notation
Quick Reference Card
| Pattern | Factored Form |
|---|---|
| GCF | Pull out common factor first |
| Difference of squares | a² - b² = (a+b)(a-b) |
| Perfect square (sum) | a² + 2ab + b² = (a+b)² |
| Perfect square (diff) | a² - 2ab + b² = (a-b)² |
| Sum of cubes | a³+b³ = (a+b)(a²-ab+b²) |
| Difference of cubes | a³-b³ = (a-b)(a²+ab+b²) |
| Factor Theorem | f(k)=0 ↔ (x-k) is a factor |
| Remainder Theorem | f(k) = remainder of f(x)÷(x-k) |
| Sum of squares | a²+b² — does NOT factor (real numbers) |