C1 — GCF Factoring: The Zeroth Step
The GCF is called the "zeroth step" because it must happen before you apply any other technique. Skipping it forces you to work with larger, messier expressions — and often leads to incomplete factoring.
Greatest Common Factor (GCF)
The GCF of a polynomial is the largest monomial that divides every term. It includes:
- The GCF of all numeric coefficients (largest integer dividing each)
- The lowest power of each variable appearing in every term
Factor it out using the distributive law in reverse: \( ac + bc = c(a+b) \).
Mini-example: Factor \( 6x^3y^2 - 9x^2y^3 + 3x^2y \).
GCF of coefficients: \( \gcd(6, 9, 3) = 3 \)
Lowest power of \( x \): \( x^2 \). Lowest power of \( y \): \( y^1 \).
GCF \( = 3x^2y \)
Result: \( 3x^2y(2xy - 3y^2 + 1) \) — check by distributing back. ✓
Stopping too early. After factoring out the GCF, always check whether the remaining factor can be factored further. For example: \( 4x^2 - 16 = 4(x^2 - 4) \). But \( x^2 - 4 = (x-2)(x+2) \), so the complete factoring is \( 4(x-2)(x+2) \). Stopping at \( 4(x^2-4) \) is not fully factored.
C2 — Factoring by Grouping
Factoring by Grouping
For a 4-term polynomial: split into two pairs, factor the GCF from each pair, then factor out the common binomial that emerges.
Standard pattern: \( ax + ay + bx + by = a(x+y) + b(x+y) = (a+b)(x+y) \)
Mini-example: Factor \( x^3 + 2x^2 - 4x - 8 \).
Group: \( (x^3 + 2x^2) + (-4x - 8) \)
Factor each group: \( x^2(x + 2) - 4(x + 2) \)
Factor the common binomial \( (x+2) \): \( (x^2 - 4)(x + 2) \)
Continue — \( x^2-4 \) is a difference of squares: \( (x-2)(x+2)(x+2) = (x-2)(x+2)^2 \)
Check: \( (x-2)(x+2)^2 = (x-2)(x^2+4x+4) = x^3+4x^2+4x-2x^2-8x-8 = x^3+2x^2-4x-8 \) ✓
The key to grouping is choosing the right pairs. The binomials inside each group's GCF must be identical after factoring — if they're not, try rearranging the terms or regrouping. The common binomial is the signal that the method worked.
C3 — Difference of Squares
Difference of Squares
For any expressions \( a \) and \( b \):
\[ a^2 - b^2 = (a - b)(a + b) \]
Proof: \( (a-b)(a+b) = a^2 + ab - ab - b^2 = a^2 - b^2 \). The middle terms cancel.
Both a and b can be any algebraic expression — they don't have to be single variables.
Examples:
- \( x^2 - 25 = x^2 - 5^2 = (x-5)(x+5) \)
- \( 4x^2 - 9 = (2x)^2 - 3^2 = (2x-3)(2x+3) \)
- \( x^4 - 16 = (x^2)^2 - 4^2 = (x^2-4)(x^2+4) = (x-2)(x+2)(x^2+4) \) — applied twice!
Sum of squares does NOT factor over ℝ. \( a^2 + b^2 \) cannot be written as a product of two real binomials. To see why: if \( a^2 + b^2 = (a + pb)(a + qb) \), then matching coefficients of \( ab \) gives \( p+q = 0 \) and of \( b^2 \) gives \( pq = 1 \). So \( q = -p \) and \( -p^2 = 1 \), which has no real solution. For calculus purposes, \( x^2 + 4 \) stays as \( x^2 + 4 \).
Don't confuse with a perfect square trinomial. \( a^2 - b^2 \) has two terms with a minus sign — that's the difference of squares pattern. The expression \( (a-b)^2 = a^2 - 2ab + b^2 \) has three terms. If you write \( x^2 - 9 = (x-3)^2 \), check: \( (x-3)^2 = x^2 - 6x + 9 \neq x^2 - 9 \). Always use both factors: \( (x-3)(x+3) \).
C4 — Quadratic Trinomials
Case 1 — Monic (leading coefficient = 1): \( x^2 + bx + c \)
Product-Sum Method
Find two integers \( p \) and \( q \) such that \( pq = c \) (product) and \( p + q = b \) (sum). Then:
\[ x^2 + bx + c = (x + p)(x + q) \]
Mini-example: Factor \( x^2 + 5x + 6 \).
Need \( pq = 6 \) and \( p+q = 5 \). Pairs for 6: \( (1,6), (2,3), (-1,-6), (-2,-3) \). Only \( (2,3) \) sums to 5.
Result: \( x^2 + 5x + 6 = (x+2)(x+3) \). Check: \( (x+2)(x+3) = x^2 + 5x + 6 \) ✓
Case 2 — Non-monic (leading coefficient ≠ 1): \( ax^2 + bx + c \)
The ac-Method
Compute \( ac \). Find two integers \( p, q \) with \( pq = ac \) and \( p + q = b \). Split the middle term \( bx \) as \( px + qx \), then factor by grouping.
Mini-example: Factor \( 2x^2 + 7x + 3 \).
\( ac = 2 \cdot 3 = 6 \). Need \( pq = 6 \) and \( p+q = 7 \): try \( (1, 6) \): \( 1 \cdot 6 = 6 \) ✓, \( 1+6=7 \) ✓.
Split: \( 2x^2 + x + 6x + 3 = x(2x+1) + 3(2x+1) = (x+3)(2x+1) \). Check ✓
Product vs. Sum confusion. In both methods, the two numbers must multiply to \( c \) (or \( ac \)) and add to \( b \). A common error is adding when you should multiply, or multiplying when you should add. Write \( pq = c \) and \( p+q = b \) explicitly on paper before hunting for the pair — the labels remind you which operation is which.
C5 — The Factor Theorem and Synthetic Division
Factor Theorem
If \( f(a) = 0 \) for some constant \( a \), then \( (x - a) \) is a factor of \( f(x) \). Conversely, if \( (x - a) \) is a factor of \( f(x) \), then \( f(a) = 0 \).
Strategy for degree ≥ 3:
- Apply the Rational Root Theorem: test values \( \pm\frac{\text{factors of constant term}}{\text{factors of leading coefficient}} \).
- When you find a root \( a \) where \( f(a) = 0 \), use synthetic division to divide by \( (x - a) \).
- Factor the quotient (now degree \( n-1 \)) using any applicable method.
- Repeat until fully factored.
Synthetic division example: divide \( x^3 - 6x^2 + 11x - 6 \) by \( (x-1) \).
Root = 1. Write coefficients: 1, −6, 11, −6.
1 | 1 -6 11 -6
| 1 -5 6
——————————————————
1 -5 6 0
Quotient: \( x^2 - 5x + 6 = (x-2)(x-3) \)
Full factoring: \( (x-1)(x-2)(x-3) \). Check by expanding. ✓
Missing placeholder zeros in synthetic division. If a polynomial has a missing term — for example, \( x^3 + 1 \) has no \( x^2 \) or \( x \) term — you must insert 0 placeholders. The coefficients are \( [1, 0, 0, 1] \). Skipping the zeros shifts every column and produces a completely wrong quotient.
C6 — Factoring with Fractional and Negative Exponents
This is the technique that appears in virtually every calculus problem. When you differentiate using the power rule, the result often contains terms with fractional or negative exponents. Setting the derivative equal to zero requires factoring it first.
The Golden Rule: Factor Out the Smallest Exponent
When an expression contains terms with fractional or negative exponents, factor out the term with the most negative (smallest) exponent.
Why: factoring out \( x^k \) (the smallest exponent) means every remaining term has exponent \( m - k \geq 0 \), producing a cleaner expression with only positive powers inside the parentheses.
Key rule: \( x^m \div x^k = x^{m-k} \) (quotient rule for exponents).
Mini-example: Factor \( 3x^{1/2} - 6x^{-1/2} \).
Smallest exponent: \( -1/2 \). Factor out \( x^{-1/2} \):
\[ 3x^{1/2} - 6x^{-1/2} = x^{-1/2}\!\left(3x^{1/2 - (-1/2)} - 6x^{-1/2 - (-1/2)}\right) = x^{-1/2}(3x^1 - 6x^0) = x^{-1/2}(3x - 6) \]
Factor out the 3: \( 3x^{-1/2}(x - 2) \), or equivalently \( \dfrac{3(x-2)}{\sqrt{x}} \)
Calculus application: if this were \( f'(x) \), solving \( f'(x) = 0 \) gives \( x^{-1/2}(3x-6) = 0 \). Since \( x^{-1/2} \neq 0 \) for \( x > 0 \), the only solution is \( 3x - 6 = 0 \), i.e., \( x = 2 \) — the critical point.
Factoring out the wrong (largest) exponent. If you factor out \( x^{1/2} \) instead of \( x^{-1/2} \) from \( 3x^{1/2} - 6x^{-1/2} \), you get \( x^{1/2}(3 - 6x^{-1}) \) — which still has a negative exponent inside and is harder to work with. Always factor out the most negative exponent to get the simplest, cleanest form.