Concept Overview
Factoring is the process of rewriting a polynomial expression as a product of simpler expressions. It is a reverse operation of polynomial multiplication and is essential for solving equations, simplifying expressions, and analyzing algebraic structure. This module explores several key techniques and special cases of factoring.
Key Vocabulary
- Factor: An expression that divides another expression exactly.
- Greatest Common Factor (GCF): The largest expression that divides all terms of a polynomial.
- Trinomial: A polynomial with three terms.
- Perfect Square: A number or expression that is the square of another.
Factoring Out the Greatest Common Factor (GCF)
Factoring out the Greatest Common Factor (GCF) is often the first step when simplifying a polynomial. It involves rewriting each term as a product of its factors, identifying what they have in common, and then expressing the original polynomial as a product using those shared factors.
One helpful strategy is to break each term into its prime factorization (including variables), then look for what appears in every term.
Prime Factorization Strategy
Factor: \( 18x^2y + 12xy^2 \)
- \( 18x^2y = 2 \cdot 3^2 \cdot x \cdot x \cdot y \)
- \( 12xy^2 = 2^2 \cdot 3 \cdot x \cdot y \cdot y \)
Common factors: \( 2 \cdot 3 \cdot x \cdot y = 6xy \)
Factored form: \( 18x^2y + 12xy^2 = 6xy(3x + 2y) \)
General Pattern
If every term contains a common factor \( a \), then:
\( ax + ay = a(x + y) \)
This is called factoring out the GCF, or reversing the distributive property.
Example 1
Factor: \( 6x + 9 \)
GCF is 3
\( 6x + 9 = 3(2x + 3) \)
Example 2
Factor: \( 12x^2y – 8xy^2 \)
GCF is \( 4xy \)
\( 12x^2y – 8xy^2 = 4xy(3x – 2y) \)
Example 3
Factor: \( 15a^3b^2 – 10a^2b + 5ab \)
GCF is \( 5ab \)
\( 15a^3b^2 – 10a^2b + 5ab = 5ab(3a^2b – 2a + 1) \)
Factoring by Grouping
Factoring by grouping is a method used when a polynomial has four terms. The idea is to group the terms into two pairs and factor out a common factor from each pair. If done carefully, the two groups will reveal a common binomial factor.
This method builds on factoring out the GCF, but applies it twice—once in each group—before identifying a final common factor. It works best when the expression can be rearranged so that the grouping creates a match.
Pattern to Recognize
\( ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y) \)
This relies on grouping the terms into two parts, factoring each part, then factoring the shared binomial.
Example 1
Factor: \( x^2 + 3x + 2x + 6 \)
Group terms: \( (x^2 + 3x) + (2x + 6) \)
Factor each group: \( x(x + 3) + 2(x + 3) \)
Factor the common binomial: \( (x + 3)(x + 2) \)
Example 2
Factor: \( 6ab + 9a + 4b + 6 \)
Group terms: \( (6ab + 9a) + (4b + 6) \)
Factor each group: \( 3a(2b + 3) + 2(2b + 3) \)
Factor the common binomial: \( (3a + 2)(2b + 3) \)
Example 3 (Rearrangement Needed)
Factor: \( ac + 3a + 2c + 6 \)
Rearrange: \( (ac + 2c) + (3a + 6) \)
Factor: \( c(a + 2) + 3(a + 2) = (c + 3)(a + 2) \)
Factoring Trinomials (Leading Coefficient 1)
A common factoring task involves rewriting a trinomial like \( x^2 + bx + c \) as a product of two binomials. When the coefficient on \( x^2 \) is 1, the process focuses on finding two numbers that multiply to \( c \) and add to \( b \).
This is often called the trial-pair method or factoring simple trinomials. It builds intuitive fluency with signs and products.
Standard Form
\( x^2 + bx + c = (x + b_1)(x + b_2) \)
This works when the leading coefficient is 1.
Find values \( b_1 \) and \( b_2 \) such that:
- \( b_1 + b_2 = b \)
- \( b_1 \cdot b_2 = c \)
Then factor directly using:
\( x^2 + bx + c = (x + b_1)(x + b_2) \)
This area model represents the expansion of the factored form \( (x + b_1)(x + b_2) \). Each tile corresponds to a term in the expanded trinomial: \( x^2 \), \( b_2x \), \( b_1x \), and \( c \). The model visually supports the grouping strategy by highlighting how the middle terms arise from different partial products.
Example 1 (Both Positive)
Factor: \( x^2 + 7x + 12 \)
Find two numbers that multiply to 12 and add to 7: \( 3 \) and \( 4 \)
\( x^2 + 7x + 12 = (x + 3)(x + 4) \)
Example 2 (Mixed Signs)
Factor: \( x^2 – x – 12 \)
Find two numbers that multiply to \(-12\) and add to \(-1\): \( -4 \) and \( 3 \)
\( x^2 – x – 12 = (x – 4)(x + 3) \)
Example 3 (Both Negative)
Factor: \( x^2 – 11x + 24 \)
Find two numbers that multiply to 24 and add to \(-11\): \( -8 \) and \( -3 \)
\( x^2 – 11x + 24 = (x – 8)(x – 3) \)
If no such pair of numbers exists (e.g., the trinomial is prime or the leading coefficient is not 1), a different method is required — such as trial-and-error, grouping, or completing the square.
Factoring Trinomials (General Case)
When the leading coefficient of a trinomial is not 1, the factoring process requires more care. One method that works reliably is called the AC Method (sometimes called the decomposition or splitting method).
Given a trinomial in the form \( ax^2 + bx + c \), multiply the leading and constant coefficients: \( A = a \cdot c \). Then, look for two numbers that multiply to \( A \) and add to \( b \). These are used to split the middle term and factor by grouping.
AC Method Summary
\( ax^2 + bx + c = ax^2 + b_1x + b_2x + c \)
Let \( b_1 \) and \( b_2 \) be numbers such that:
- \( b_1 + b_2 = b \)
- \( b_1 \cdot b_2 = a \cdot c \)
\( b_1 \) and \( b_2 \) are used to split the middle term so we can factor by grouping:
\( ax^2 + b_1x + b_2x + c \rightarrow \text{group and factor} \)
This area model represents the expansion of the factored form \( (ax + b_1)(x + b_2) \). Each tile corresponds to a term in the expanded trinomial: \( ax^2 \), \( ab_2x \), \( b_1x \), and \( c \). The model visually supports the AC method and grouping strategy by highlighting how the middle terms arise from different partial products.
Example 1
Factor: \( 6x^2 + 11x + 4 \)
- AC = \( 6 \cdot 4 = 24 \)
- Find two numbers that multiply to 24 and add to 11: \( 3 \) and \( 8 \)
- Rewrite: \( 6x^2 + 3x + 8x + 4 \)
- Group: \( (6x^2 + 3x) + (8x + 4) \)
- Factor: \( 3x(2x + 1) + 4(2x + 1) = (3x + 4)(2x + 1) \)
Example 2
Factor: \( 2x^2 – 5x – 12 \)
- AC = \( 2 \cdot -12 = -24 \)
- Two numbers that multiply to -24 and add to -5: \( -8 \) and \( 3 \)
- Rewrite: \( 2x^2 – 8x + 3x – 12 \)
- Group: \( (2x^2 – 8x) + (3x – 12) \)
- Factor: \( 2x(x – 4) + 3(x – 4) = (2x + 3)(x – 4) \)
Visualizing the FOIL Method with an Area Model: This animation illustrates how the product of two binomials, \( (ax + b_1)(x + b_2) \), expands into four terms using the FOIL method: First \( ax \cdot x = ax^2 \), Outer \( ax \cdot b_2 = ab_2x \), Inner \( b_1 \cdot x = b_1x \), and Last \( b_1 \cdot b_2 = b_1b_2 \). Each region in the area model represents one of these partial products, reinforcing the structure of the expression: \( ax^2 + bx + c \).
Helpful Tips When Factoring
- Watch the sign of each term: The values of \( b_1 \) and \( b_2 \) may be negative—make sure the signs carry through both the area products and the binomial factors.
- Grouping can be rearranged: If factoring by grouping doesn’t work immediately, try swapping the middle terms (e.g., write \( b_2x + b_1x \) instead of \( b_1x + b_2x \)). The area model stays valid, but grouping may become easier.
- The order of binomials doesn’t matter: \( (ax + b_1)(x + b_2) \) and \( (x + b_2)(ax + b_1) \) both produce the same expanded result due to the commutative property of multiplication.
- Not all quadratics factor nicely: If you can’t find integers for \( b_1 \) and \( b_2 \) that satisfy both conditions, the trinomial may not be factorable using rational numbers.
- The area model can be reversed: If you’re given a factored expression, you can build the rectangle to visualize the expansion and confirm your answer.
Try It Yourself!
- Factor out the GCF: \( 18x^2y + 24xy^2 \)
- Factor by grouping: \( x^2 + 2x + 3x + 6 \)
- Factor the trinomial: \( x^2 + 7x + 12 \)
- Factor the trinomial: \( 3x^2 + 10x + 8 \)
- Factor completely: \( 4x^2 – 20x + 25 \)
Reveal Answers
- \( 6xy(3x + 4y) \)
- \( (x^2 + 2x) + (3x + 6) = x(x + 2) + 3(x + 2) = (x + 2)(x + 3) \)
- \( (x + 3)(x + 4) \)
- \( (3x + 4)(x + 2) \)
- \( (2x – 5)^2 \)