Study guide
4.2 Simplifying, Combining Like Terms, & Factoring
ACCUPLACER Quantitative Reasoning, Algebra, and Statistics Exam
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1 Chapter 1: Rational Numbers & Operations
1.1 Rational Number Arithmetic & Signed Numbers 1.2 Fractions, Decimals, & Mixed Numbers Operations 1.3 Absolute Value, Number Line, & Estimation
2 Chapter 2: Ratios, Proportions & Rates
2.1 Ratios, Unit Rates, & Dimensional Analysis 2.2 Proportions & Direct/Inverse Variation 2.3 Percent Applications, Markup, Discount, & Simple Interest
3 Chapter 3: Exponents & Scientific Notation
3.1 Integer Exponents & Laws of Exponents 3.2 Scientific Notation & Operations
4 Chapter 4: Algebraic Expressions
4.1 Evaluating & Translating Algebraic Expressions 4.2 Simplifying, Combining Like Terms, & Factoring
5 Chapter 5: Linear Equations & Inequalities
5.1 Solving Single-Variable Linear Equations 5.2 Solving & Graphing Linear Inequalities 5.3 Systems of Linear Equations
6 Chapter 6: Linear Applications & Graphing
6.1 Slope & Rate of Change 6.2 Forms of Linear Equations & Graphing 6.3 Linear Modeling & Inequality Graphs
7 Chapter 7: Probability & Set Theory
7.1 Fundamental Probability Concepts 7.2 Set Theory, Operations, & Venn Diagrams
8 Chapter 8: Descriptive Statistics & Data Displays
8.1 Measures of Center: Mean, Median, Mode, & Weighted Average 8.2 Measures of Spread: Range, IQR, Variance, & Standard Deviation 8.3 Data Displays: Histograms, Box Plots, & Scatter Plots
9 Chapter 9: Geometry Concepts & Measurement
9.1 2D Geometry: Perimeter, Area, & Circles 9.2 3D Geometry: Surface Area & Volume of Solids 9.3 Triangles, Angles, & the Pythagorean Theorem
4.2 Simplifying, Combining Like Terms, & Factoring
Key Takeaways Explain This Topic AI
Like terms are algebraic terms possessing the exact same variable bases raised to identical exponents; only their numerical coefficients are combined through addition or subtraction.
The Distributive Property expands products across additions and subtractions: a(b + c) = ab + ac; distributing a negative coefficient reverses the sign of every term inside the parentheses.
Polynomial multiplication relies on systematic term distribution; for binomials, the FOIL method yields (a + b)(c + d) = ac + ad + bc + bd.
Factoring reverses polynomial expansion through a hierarchical sequence of strategies: Greatest Common Factor (GCF), Factoring by Grouping (4 terms), Quadratic Trinomial Factoring (ac-method), and Difference of Two Squares (a^2 - b^2 = (a - b)(a + b)).
Simplifying rational algebraic expressions requires completely factoring the numerator and denominator into prime factors and canceling common non-zero factors; terms separated by addition or subtraction cannot be canceled individually.
Last updated: August 2026
Combining Like Terms & Algebraic Structure
Simplifying algebraic expressions is the core mechanical skill underpinning algebra. An algebraic expression is considered to be in simplest form when all grouping symbols have been removed and all like terms have been combined.
Defining Like Terms vs. Unlike Terms
Like terms are terms that contain the exact same variables raised to the exact same exponents . The numerical coefficients of like terms can be different, but the variable factors must match identically. Like Terms: 7 x 2 y 3 and − 4 x 2 y 3 ⟹ ( 7 − 4 ) x 2 y 3 = 3 x 2 y 3 \text{Like Terms: } 7x^2 y^3 \quad \text{and} \quad -4x^2 y^3 \implies (7 - 4)x^2 y^3 = 3x^2 y^3 Like Terms: 7 x 2 y 3 and − 4 x 2 y 3 ⟹ ( 7 − 4 ) x 2 y 3 = 3 x 2 y 3
Unlike terms differ in either their variable letters or the exponents attached to those variables. Unlike terms cannot be combined into a single term through addition or subtraction. Unlike Terms: 5 x 2 and 5 x ⟹ 5 x 2 + 5 x ( Cannot be combined into 10 x 3 or 10 x 2 ) \text{Unlike Terms: } 5x^2 \quad \text{and} \quad 5x \implies 5x^2 + 5x \quad (\text{Cannot be combined into } 10x^3 \text{ or } 10x^2) Unlike Terms: 5 x 2 and 5 x ⟹ 5 x 2 + 5 x ( Cannot be combined into 10 x 3 or 10 x 2 )
Comparison Table: Like vs. Unlike Terms
Term Pair Like or Unlike? Reason Simplified Combination
8 a 2 b 8a^2 b 8 a 2 b and − 3 a 2 b -3a^2 b − 3 a 2 b Like Identical variable parts ( a 2 b a^2 b a 2 b ) 8 a 2 b + ( − 3 a 2 b ) = 5 a 2 b 8a^2 b + (-3a^2 b) = \mathbf{5a^2 b} 8 a 2 b + ( − 3 a 2 b ) = 5 a 2 b
6 x y 2 6xy^2 6 x y 2 and 6 x 2 y 6x^2 y 6 x 2 y Unlike Exponents on x x x and y y y are reversed 6 x y 2 + 6 x 2 y 6xy^2 + 6x^2 y 6 x y 2 + 6 x 2 y (Already simplified)
9 x 9\sqrt{x} 9 x and − 2 x -2\sqrt{x} − 2 x Like Identical radical base ( x 1 / 2 x^{1/2} x 1/2 ) 9 x − 2 x = 7 x 9\sqrt{x} - 2\sqrt{x} = \mathbf{7\sqrt{x}} 9 x − 2 x = 7 x
4 m 3 4m^3 4 m 3 and 4 m 4 4m^4 4 m 4 Unlike Different exponents ( 3 3 3 vs. 4 4 4 ) 4 m 3 + 4 m 4 4m^3 + 4m^4 4 m 3 + 4 m 4 (Already simplified)
− 11 -11 − 11 and 25 25 25 Like Constant numbers (degree 0 0 0 ) − 11 + 25 = 14 -11 + 25 = \mathbf{14} − 11 + 25 = 14
5 a b 5ab 5 ab and − 2 b a -2ba − 2 ba Like Commutative property of multiplication ( a b = b a ab = ba ab = ba ) 5 a b − 2 b a = 3 a b 5ab - 2ba = \mathbf{3ab} 5 ab − 2 ba = 3ab
The Distributive Property & Managing Signed Numbers
The Distributive Property allows a multiplier outside parentheses to be distributed to every term inside the parentheses: a ( b + c ) = a b + a c a(b + c) = ab + ac a ( b + c ) = ab + a c a ( b − c ) = a b − a c a(b - c) = ab - ac a ( b − c ) = ab − a c
Distributing Negative Signs and Coefficients
When distributing a negative number or a negative sign across parentheses, the sign of every single term inside the grouping symbol must be reversed: − ( a x 2 − b x + c ) = − a x 2 + b x − c -(ax^2 - bx + c) = -ax^2 + bx - c − ( a x 2 − b x + c ) = − a x 2 + b x − c − 4 ( 2 x 2 − 5 x + 3 ) = − 4 ( 2 x 2 ) − 4 ( − 5 x ) − 4 ( 3 ) = − 8 x 2 + 20 x − 12 -4(2x^2 - 5x + 3) = -4(2x^2) - 4(-5x) - 4(3) = -8x^2 + 20x - 12 − 4 ( 2 x 2 − 5 x + 3 ) = − 4 ( 2 x 2 ) − 4 ( − 5 x ) − 4 ( 3 ) = − 8 x 2 + 20 x − 12
Simplifying Expressions with Nested Grouping Symbols
When simplifying expressions containing parentheses ( ) ( ) ( ) , brackets [ ] [ ] [ ] , or braces { } \{ \} { } , work systematically from the innermost grouping outward: Expression: 5 − 2 [ 3 x − 4 ( x − 2 ) + 7 ] \text{Expression: } 5 - 2[3x - 4(x - 2) + 7] Expression: 5 − 2 [ 3 x − 4 ( x − 2 ) + 7 ]
Distribute innermost multiplier − 4 -4 − 4 : 5 − 2 [ 3 x − 4 x + 8 + 7 ] 5 - 2[3x - 4x + 8 + 7] 5 − 2 [ 3 x − 4 x + 8 + 7 ]
- Combine like terms inside brackets : 5 − 2 [ − x + 15 ] 5 - 2[-x + 15] 5 − 2 [ − x + 15 ]
- Distribute outer multiplier − 2 -2 − 2 : 5 + 2 x − 30 5 + 2x - 30 5 + 2 x − 30
- Combine constants : 2 x − 25 2x - 25 2 x − 25
- Multiplying Polynomials
Multiplying polynomials requires applying the distributive property repeatedly along with the Product Rule of Exponents ( x m ⋅ x n = x m + n x^m \cdot x^n = x^{m+n} x m ⋅ x n = x m + n ).
1. Multiplying a Monomial by a Polynomial
Distribute the monomial to every term in the polynomial, multiplying numerical coefficients and adding variable exponents: − 3 x 2 y ( 4 x 3 − 5 x y 2 + 2 y 3 ) = ( − 3 x 2 y ) ( 4 x 3 ) + ( − 3 x 2 y ) ( − 5 x y 2 ) + ( − 3 x 2 y ) ( 2 y 3 ) -3x^2 y (4x^3 - 5xy^2 + 2y^3) = (-3x^2 y)(4x^3) + (-3x^2 y)(-5xy^2) + (-3x^2 y)(2y^3) − 3 x 2 y ( 4 x 3 − 5 x y 2 + 2 y 3 ) = ( − 3 x 2 y ) ( 4 x 3 ) + ( − 3 x 2 y ) ( − 5 x y 2 ) + ( − 3 x 2 y ) ( 2 y 3 ) = − 12 x 5 y + 15 x 3 y 3 − 6 x 2 y 4 = -12x^5 y + 15x^3 y^3 - 6x^2 y^4 = − 12 x 5 y + 15 x 3 y 3 − 6 x 2 y 4
2. Multiplying Two Binomials (The FOIL Method)
For two binomials ( a + b ) ( c + d ) (a + b)(c + d) ( a + b ) ( c + d ) , the FOIL acronym ensures all four pairwise products are calculated: ( a + b ) ( c + d ) = a c ⏟ First + a d ⏟ Outer + b c ⏟ Inner + b d ⏟ Last (a + b)(c + d) = \underbrace{ac}_{\text{First}} + \underbrace{ad}_{\text{Outer}} + \underbrace{bc}_{\text{Inner}} + \underbrace{bd}_{\text{Last}} ( a + b ) ( c + d ) = First a c + Outer a d + Inner b c + Last b d
- Worked Example : ( 3 x − 4 ) ( 2 x + 5 ) (3x - 4)(2x + 5) ( 3 x − 4 ) ( 2 x + 5 )
- First : ( 3 x ) ( 2 x ) = 6 x 2 (3x)(2x) = 6x^2 ( 3 x ) ( 2 x ) = 6 x 2
- Outer : ( 3 x ) ( 5 ) = + 15 x (3x)(5) = +15x ( 3 x ) ( 5 ) = + 15 x
- Inner : ( − 4 ) ( 2 x ) = − 8 x (-4)(2x) = -8x ( − 4 ) ( 2 x ) = − 8 x
- Last : ( − 4 ) ( 5 ) = − 20 (-4)(5) = -20 ( − 4 ) ( 5 ) = − 20
Combine middle like terms : 6 x 2 + ( 15 x − 8 x ) − 20 = 6 x 2 + 7 x − 20 6x^2 + (15x - 8x) - 20 = \mathbf{6x^2 + 7x - 20} 6 x 2 + ( 15 x − 8 x ) − 20 = 6 x 2 + 7x − 20
3. Special Polynomial Product Patterns
Mastering special product formulas allows instant expansion and forms the foundation for rapid factoring:
Pattern Name Formula Worked Example
Difference of Squares ( a − b ) ( a + b ) = a 2 − b 2 (a - b)(a + b) = a^2 - b^2 ( a − b ) ( a + b ) = a 2 − b 2 ( 4 x − 7 ) ( 4 x + 7 ) = ( 4 x ) 2 − ( 7 ) 2 = 16 x 2 − 49 (4x - 7)(4x + 7) = (4x)^2 - (7)^2 = \mathbf{16x^2 - 49} ( 4 x − 7 ) ( 4 x + 7 ) = ( 4 x ) 2 − ( 7 ) 2 = 16 x 2 − 49
Square of a Sum ( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2 ( 3 x + 5 ) 2 = ( 3 x ) 2 + 2 ( 3 x ) ( 5 ) + 5 2 = 9 x 2 + 30 x + 25 (3x + 5)^2 = (3x)^2 + 2(3x)(5) + 5^2 = \mathbf{9x^2 + 30x + 25} ( 3 x + 5 ) 2 = ( 3 x ) 2 + 2 ( 3 x ) ( 5 ) + 5 2 = 9 x 2 + 30x + 25
Square of a Difference ( a − b ) 2 = a 2 − 2 a b + b 2 (a - b)^2 = a^2 - 2ab + b^2 ( a − b ) 2 = a 2 − 2 ab + b 2 ( 2 x − 3 y ) 2 = ( 2 x ) 2 − 2 ( 2 x ) ( 3 y ) + ( 3 y ) 2 = 4 x 2 − 12 x y + 9 y 2 (2x - 3y)^2 = (2x)^2 - 2(2x)(3y) + (3y)^2 = \mathbf{4x^2 - 12xy + 9y^2} ( 2 x − 3 y ) 2 = ( 2 x ) 2 − 2 ( 2 x ) ( 3 y ) + ( 3 y ) 2 = 4 x 2 − 12xy + 9 y 2
The Classic ACCUPLACER Trap : Exponents do never distribute across addition: ( a + b ) 2 ≠ a 2 + b 2 (a + b)^2 \neq a^2 + b^2 ( a + b ) 2 = a 2 + b 2 . Always include the middle cross-term 2 a b 2ab 2 ab .
4. Multiplying Higher-Degree Polynomials (Binomial × \times × Trinomial)
To multiply a binomial by a trinomial, distribute each term of the binomial across all three terms of the trinomial: ( 2 x − 3 ) ( x 2 + 4 x − 5 ) = 2 x ( x 2 + 4 x − 5 ) − 3 ( x 2 + 4 x − 5 ) (2x - 3)(x^2 + 4x - 5) = 2x(x^2 + 4x - 5) - 3(x^2 + 4x - 5) ( 2 x − 3 ) ( x 2 + 4 x − 5 ) = 2 x ( x 2 + 4 x − 5 ) − 3 ( x 2 + 4 x − 5 ) = ( 2 x 3 + 8 x 2 − 10 x ) − ( 3 x 2 + 12 x − 15 ) = (2x^3 + 8x^2 - 10x) - (3x^2 + 12x - 15) = ( 2 x 3 + 8 x 2 − 10 x ) − ( 3 x 2 + 12 x − 15 ) = 2 x 3 + ( 8 x 2 − 3 x 2 ) + ( − 10 x − 12 x ) + 15 = 2x^3 + (8x^2 - 3x^2) + (-10x - 12x) + 15 = 2 x 3 + ( 8 x 2 − 3 x 2 ) + ( − 10 x − 12 x ) + 15 = 2 x 3 + 5 x 2 − 22 x + 15 = \mathbf{2x^3 + 5x^2 - 22x + 15} = 2 x 3 + 5 x 2 − 22x + 15
Factoring Algebraic Expressions: The Complete Hierarchy
Factoring is the inverse operation of polynomial multiplication. It decomposes a polynomial into a product of simpler irreducible factors. On the ACCUPLACER QAS exam, you must follow a systematic multi-step factoring algorithm. Step 1: Always Factor Out the Greatest Common Factor (GCF) First.
Step 2: Count the Number of Remaining Terms:
- 2 Terms: Check for Difference of Two Squares (a^2 - b^2 = (a-b)(a+b)).
- 3 Terms: Factor as a Quadratic Trinomial (x^2 + bx + c or ac-Method for ax^2 + bx + c).
- 4 Terms: Factor by Grouping in Pairs.
Step 3: Verify That Every Factor is Fully Reduced (Prime).
Strategy 1: Factoring Out the Greatest Common Factor (GCF)
The GCF is the largest monomial that divides evenly into every term of the polynomial. Its coefficient is the numerical GCF, and its variable part contains the lowest exponent of each shared variable.
Worked Example : Factor: 12 x 4 y 3 − 18 x 3 y 5 + 6 x 2 y 2 \text{Factor: } 12x^4 y^3 - 18x^3 y^5 + 6x^2 y^2 Factor: 12 x 4 y 3 − 18 x 3 y 5 + 6 x 2 y 2
Numerical GCF of 12 , 18 , 6 12, 18, 6 12 , 18 , 6 is 6 6 6 .
Lowest power of x x x is x 2 x^2 x 2 .
Lowest power of y y y is y 2 y^2 y 2 .
Overall GCF = 6 x 2 y 2 \text{GCF} = 6x^2 y^2 GCF = 6 x 2 y 2 .
Divide each term by the GCF: 12 x 4 y 3 6 x 2 y 2 = 2 x 2 y , − 18 x 3 y 5 6 x 2 y 2 = − 3 x y 3 , 6 x 2 y 2 6 x 2 y 2 = + 1 \frac{12x^4 y^3}{6x^2 y^2} = 2x^2 y, \quad \frac{-18x^3 y^5}{6x^2 y^2} = -3xy^3, \quad \frac{6x^2 y^2}{6x^2 y^2} = +1 6 x 2 y 2 12 x 4 y 3 = 2 x 2 y , 6 x 2 y 2 − 18 x 3 y 5 = − 3 x y 3 , 6 x 2 y 2 6 x 2 y 2 = + 1
Factored form: 6 x 2 y 2 ( 2 x 2 y − 3 x y 3 + 1 ) \mathbf{6x^2 y^2 (2x^2 y - 3xy^3 + 1)} 6 x 2 y 2 ( 2 x 2 y − 3x y 3 + 1 )
Strategy 2: Factoring by Grouping (4 Terms)
When a polynomial has four terms, group them into pairs, factor the GCF from each pair, and extract the common binomial factor.
Worked Example : Factor: 3 x 3 − 6 x 2 + 4 x − 8 \text{Factor: } 3x^3 - 6x^2 + 4x - 8 Factor: 3 x 3 − 6 x 2 + 4 x − 8
- Group into pairs: ( 3 x 3 − 6 x 2 ) + ( 4 x − 8 ) (3x^3 - 6x^2) + (4x - 8) ( 3 x 3 − 6 x 2 ) + ( 4 x − 8 )
- Factor GCF from first pair: 3 x 2 ( x − 2 ) 3x^2(x - 2) 3 x 2 ( x − 2 )
- Factor GCF from second pair: + 4 ( x − 2 ) +4(x - 2) + 4 ( x − 2 )
Factor out the common binomial ( x − 2 ) (x - 2) ( x − 2 ) : ( x − 2 ) ( 3 x 2 + 4 ) \mathbf{(x - 2)(3x^2 + 4)} ( x − 2 ) ( 3 x 2 + 4 )
Strategy 3: Factoring Quadratic Trinomials ( x 2 + b x + c x^2 + bx + c x 2 + b x + c , Leading Coefficient a = 1 a = 1 a = 1 )
To factor x 2 + b x + c x^2 + bx + c x 2 + b x + c , find two integers p p p and q q q that satisfy two simultaneous conditions:
- Product : p ⋅ q = c p \cdot q = c p ⋅ q = c
- Sum : p + q = b p + q = b p + q = b
Once found, the factored form is ( x + p ) ( x + q ) (x + p)(x + q) ( x + p ) ( x + q ) .
Sign Rules for Trinomial Factoring:
If c > 0 c > 0 c > 0 and b > 0 b > 0 b > 0 : Both p p p and q q q are positive ( + , + + , + + , + ). Example: x 2 + 8 x + 12 = ( x + 6 ) ( x + 2 ) x^2 + 8x + 12 = (x + 6)(x + 2) x 2 + 8 x + 12 = ( x + 6 ) ( x + 2 )
If c > 0 c > 0 c > 0 and b < 0 b < 0 b < 0 : Both p p p and q q q are negative ( − , − - , - − , − ). Example: x 2 − 9 x + 20 = ( x − 4 ) ( x − 5 ) x^2 - 9x + 20 = (x - 4)(x - 5) x 2 − 9 x + 20 = ( x − 4 ) ( x − 5 )
If c < 0 c < 0 c < 0 : One factor is positive and one is negative ( + , − + , - + , − ), with the larger absolute value matching the sign of b b b .
Example ( b > 0 b > 0 b > 0 ): x 2 + 2 x − 15 = ( x + 5 ) ( x − 3 ) x^2 + 2x - 15 = (x + 5)(x - 3) x 2 + 2 x − 15 = ( x + 5 ) ( x − 3 )
Example ( b < 0 b < 0 b < 0 ): x 2 − 5 x − 24 = ( x − 8 ) ( x + 3 ) x^2 - 5x - 24 = (x - 8)(x + 3) x 2 − 5 x − 24 = ( x − 8 ) ( x + 3 )
Strategy 4: Factoring Quadratic Trinomials ( a x 2 + b x + c ax^2 + bx + c a x 2 + b x + c , Leading Coefficient a ≠ 1 a \neq 1 a = 1 )
When a ≠ 1 a \neq 1 a = 1 and cannot be factored out via GCF, use the a c ac a c -Method (Master Product Method) :
Multiply a a a and c c c to obtain the target product a c ac a c .
Find two integers p p p and q q q whose product is a c ac a c and whose sum is b b b .
Split the middle linear term b x bx b x into p x + q x px + qx p x + q x .
Factor the resulting four-term polynomial by grouping.
Worked Example : Factor: 6 x 2 + 11 x − 10 \text{Factor: } 6x^2 + 11x - 10 Factor: 6 x 2 + 11 x − 10
Target product: a c = 6 ⋅ ( − 10 ) = − 60 ac = 6 \cdot (-10) = -60 a c = 6 ⋅ ( − 10 ) = − 60 .
Target sum: b = + 11 b = +11 b = + 11 .
Factor pairs of − 60 -60 − 60 : Factors + 15 +15 + 15 and − 4 -4 − 4 satisfy 15 ⋅ ( − 4 ) = − 60 15 \cdot (-4) = -60 15 ⋅ ( − 4 ) = − 60 and 15 + ( − 4 ) = 11 15 + (-4) = 11 15 + ( − 4 ) = 11 .
- Rewrite middle term: 6 x 2 + 15 x − 4 x − 10 6x^2 + 15x - 4x - 10 6 x 2 + 15 x − 4 x − 10
- Factor by grouping: 3 x ( 2 x + 5 ) − 2 ( 2 x + 5 ) 3x(2x + 5) - 2(2x + 5) 3 x ( 2 x + 5 ) − 2 ( 2 x + 5 )
- Extract common binomial: ( 2 x + 5 ) ( 3 x − 2 ) \mathbf{(2x + 5)(3x - 2)} ( 2x + 5 ) ( 3x − 2 )
- Strategy 5: Factoring Difference of Two Squares ( a 2 − b 2 a^2 - b^2 a 2 − b 2 )
Any binomial consisting of two perfect squares separated by a minus sign factors into conjugate binomials: a 2 − b 2 = ( a − b ) ( a + b ) a^2 - b^2 = (a - b)(a + b) a 2 − b 2 = ( a − b ) ( a + b )
Worked Examples :
25 x 2 − 49 = ( 5 x ) 2 − ( 7 ) 2 = ( 5 x − 7 ) ( 5 x + 7 ) 25x^2 - 49 = (5x)^2 - (7)^2 = \mathbf{(5x - 7)(5x + 7)} 25 x 2 − 49 = ( 5 x ) 2 − ( 7 ) 2 = ( 5x − 7 ) ( 5x + 7 )
Multi-Stage Factoring : x 4 − 81 = ( x 2 − 9 ) ( x 2 + 9 ) = ( x − 3 ) ( x + 3 ) ( x 2 + 9 ) x^4 - 81 = (x^2 - 9)(x^2 + 9) = \mathbf{(x - 3)(x + 3)(x^2 + 9)} x 4 − 81 = ( x 2 − 9 ) ( x 2 + 9 ) = ( x − 3 ) ( x + 3 ) ( x 2 + 9 )
Important Note : The Sum of Squares ( a 2 + b 2 a^2 + b^2 a 2 + b 2 ) is prime and cannot be factored over real numbers.
Simplifying Rational Algebraic Expressions
A rational algebraic expression is a fraction whose numerator and denominator are polynomials. Rational expressions are simplified using the Fundamental Principle of Fractions: P ⋅ R Q ⋅ R = P Q ( Q ≠ 0 , R ≠ 0 ) \frac{P \cdot R}{Q \cdot R} = \frac{P}{Q} \quad (Q \neq 0, R \neq 0) Q ⋅ R P ⋅ R = Q P ( Q = 0 , R = 0 )
The Cardinal Rule: Factor Completely Before Canceling
NEVER cancel individual terms across addition or subtraction. You may only cancel entire multiplicative factors that appear in both the numerator and denominator. INCORRECT: x 2 − 9 x + 3 ≠ x − 3 (by canceling x 2 / x and − 9 / 3 ) \text{INCORRECT: } \frac{x^2 - 9}{x + 3} \neq x - 3 \text{ (by canceling } x^2/x \text{ and } -9/3) INCORRECT: x + 3 x 2 − 9 = x − 3 (by canceling x 2 / x and − 9/3 ) CORRECT: x 2 − 9 x + 3 = ( x − 3 ) ( x + 3 ) x + 3 = x − 3 ( x ≠ − 3 ) \text{CORRECT: } \frac{x^2 - 9}{x + 3} = \frac{(x - 3)\cancel{(x + 3)}}{\cancel{x + 3}} = \mathbf{x - 3} \quad (x \neq -3) CORRECT: x + 3 x 2 − 9 = x + 3 ( x − 3 ) ( x + 3 ) = x − 3 ( x = − 3 )
Managing Opposite Factors (The − 1 -1 − 1 Rule)
When a factor in the numerator is the exact negative of a factor in the denominator ( a − b a - b a − b versus b − a b - a b − a ), factor out − 1 -1 − 1 : b − a = − ( a − b ) ⟹ a − b b − a = a − b − ( a − b ) = − 1 b - a = -(a - b) \implies \frac{a - b}{b - a} = \frac{a - b}{-(a - b)} = \mathbf{-1} b − a = − ( a − b ) ⟹ b − a a − b = − ( a − b ) a − b = − 1
Worked Example : 2 x − 6 9 − x 2 = 2 ( x − 3 ) ( 3 − x ) ( 3 + x ) = 2 ( x − 3 ) − ( x − 3 ) ( 3 + x ) = − 2 x + 3 \frac{2x - 6}{9 - x^2} = \frac{2(x - 3)}{(3 - x)(3 + x)} = \frac{2(x - 3)}{-(x - 3)(3 + x)} = \mathbf{-\frac{2}{x + 3}} 9 − x 2 2 x − 6 = ( 3 − x ) ( 3 + x ) 2 ( x − 3 ) = − ( x − 3 ) ( 3 + x ) 2 ( x − 3 ) = − x + 3 2
- Comprehensive Step-by-Step Worked Examples
- Worked Example 1: Multi-Step Expansion and Nested Simplification
Simplify the expression completely: E = 3 ( x − 4 ) 2 − 2 ( x + 2 ) ( x − 5 ) − ( x 2 − 14 x + 8 ) E = 3(x - 4)^2 - 2(x + 2)(x - 5) - (x^2 - 14x + 8) E = 3 ( x − 4 ) 2 − 2 ( x + 2 ) ( x − 5 ) − ( x 2 − 14 x + 8 )
Step-by-Step Solution :
Expand the squared binomial : ( x − 4 ) 2 = x 2 − 8 x + 16 (x - 4)^2 = x^2 - 8x + 16 ( x − 4 ) 2 = x 2 − 8 x + 16 3 ( x − 4 ) 2 = 3 ( x 2 − 8 x + 16 ) = 3 x 2 − 24 x + 48 3(x - 4)^2 = 3(x^2 - 8x + 16) = 3x^2 - 24x + 48 3 ( x − 4 ) 2 = 3 ( x 2 − 8 x + 16 ) = 3 x 2 − 24 x + 48
Expand the binomial product with FOIL : ( x + 2 ) ( x − 5 ) = x 2 − 5 x + 2 x − 10 = x 2 − 3 x − 10 (x + 2)(x - 5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10 ( x + 2 ) ( x − 5 ) = x 2 − 5 x + 2 x − 10 = x 2 − 3 x − 10 − 2 ( x + 2 ) ( x − 5 ) = − 2 ( x 2 − 3 x − 10 ) = − 2 x 2 + 6 x + 20 -2(x + 2)(x - 5) = -2(x^2 - 3x - 10) = -2x^2 + 6x + 20 − 2 ( x + 2 ) ( x − 5 ) = − 2 ( x 2 − 3 x − 10 ) = − 2 x 2 + 6 x + 20
Distribute the negative sign across the last polynomial : − ( x 2 − 14 x + 8 ) = − x 2 + 14 x − 8 -(x^2 - 14x + 8) = -x^2 + 14x - 8 − ( x 2 − 14 x + 8 ) = − x 2 + 14 x − 8
Combine all components : E = ( 3 x 2 − 24 x + 48 ) + ( − 2 x 2 + 6 x + 20 ) + ( − x 2 + 14 x − 8 ) E = (3x^2 - 24x + 48) + (-2x^2 + 6x + 20) + (-x^2 + 14x - 8) E = ( 3 x 2 − 24 x + 48 ) + ( − 2 x 2 + 6 x + 20 ) + ( − x 2 + 14 x − 8 )
x 2 x^2 x 2 terms: 3 x 2 − 2 x 2 − x 2 = 0 x 2 = 0 3x^2 - 2x^2 - x^2 = 0x^2 = 0 3 x 2 − 2 x 2 − x 2 = 0 x 2 = 0
- x x x terms: − 24 x + 6 x + 14 x = − 4 x -24x + 6x + 14x = -4x − 24 x + 6 x + 14 x = − 4 x
- Constant terms: 48 + 20 − 8 = 60 48 + 20 - 8 = 60 48 + 20 − 8 = 60
- Final Result : − 4 x + 60 \mathbf{-4x + 60} − 4x + 60
- Worked Example 2: Complete Factoring of a Multi-Variable Expression
- Factor the polynomial completely over the real numbers: P = 2 x 4 y − 32 y P = 2x^4 y - 32y P = 2 x 4 y − 32 y
Step-by-Step Solution :
Factor out the GCF ( 2 y 2y 2 y ) : P = 2 y ( x 4 − 16 ) P = 2y(x^4 - 16) P = 2 y ( x 4 − 16 )
Factor x 4 − 16 x^4 - 16 x 4 − 16 as a Difference of Two Squares : x 4 − 16 = ( x 2 ) 2 − 4 2 = ( x 2 − 4 ) ( x 2 + 4 ) x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4) x 4 − 16 = ( x 2 ) 2 − 4 2 = ( x 2 − 4 ) ( x 2 + 4 ) P = 2 y ( x 2 − 4 ) ( x 2 + 4 ) P = 2y(x^2 - 4)(x^2 + 4) P = 2 y ( x 2 − 4 ) ( x 2 + 4 )
Factor the remaining Difference of Squares ( x 2 − 4 x^2 - 4 x 2 − 4 ) : x 2 − 4 = ( x − 2 ) ( x + 2 ) x^2 - 4 = (x - 2)(x + 2) x 2 − 4 = ( x − 2 ) ( x + 2 ) P = 2 y ( x − 2 ) ( x + 2 ) ( x 2 + 4 ) P = 2y(x - 2)(x + 2)(x^2 + 4) P = 2 y ( x − 2 ) ( x + 2 ) ( x 2 + 4 )
Check the sum of squares factor : ( x 2 + 4 ) (x^2 + 4) ( x 2 + 4 ) is prime over real numbers.
Final Factored Form : 2 y ( x − 2 ) ( x + 2 ) ( x 2 + 4 ) \mathbf{2y(x - 2)(x + 2)(x^2 + 4)} 2y ( x − 2 ) ( x + 2 ) ( x 2 + 4 )
Worked Example 3: Simplifying a Rational Expression with a c ac a c -Factoring and Sign Inversion
Simplify the rational expression completely and state all domain restrictions: R ( x ) = 2 x 2 + x − 15 25 − 4 x 2 R(x) = \frac{2x^2 + x - 15}{25 - 4x^2} R ( x ) = 25 − 4 x 2 2 x 2 + x − 15
Step-by-Step Solution :
Factor the numerator 2 x 2 + x − 15 2x^2 + x - 15 2 x 2 + x − 15 ( a c = 2 ( − 15 ) = − 30 ac = 2(-15) = -30 a c = 2 ( − 15 ) = − 30 , sum = + 1 = +1 = + 1 ) :
Factors are + 6 +6 + 6 and − 5 -5 − 5 : 2 x 2 + 6 x − 5 x − 15 = 2 x ( x + 3 ) − 5 ( x + 3 ) = ( 2 x − 5 ) ( x + 3 ) 2x^2 + 6x - 5x - 15 = 2x(x + 3) - 5(x + 3) = (2x - 5)(x + 3) 2 x 2 + 6 x − 5 x − 15 = 2 x ( x + 3 ) − 5 ( x + 3 ) = ( 2 x − 5 ) ( x + 3 )
Factor the denominator 25 − 4 x 2 25 - 4x^2 25 − 4 x 2 (Difference of Squares) : 25 − 4 x 2 = 5 2 − ( 2 x ) 2 = ( 5 − 2 x ) ( 5 + 2 x ) 25 - 4x^2 = 5^2 - (2x)^2 = (5 - 2x)(5 + 2x) 25 − 4 x 2 = 5 2 − ( 2 x ) 2 = ( 5 − 2 x ) ( 5 + 2 x )
Identify opposite binomial factors :
Notice that ( 5 − 2 x ) = − ( 2 x − 5 ) (5 - 2x) = -(2x - 5) ( 5 − 2 x ) = − ( 2 x − 5 ) . Rewrite denominator: ( 5 − 2 x ) ( 5 + 2 x ) = − ( 2 x − 5 ) ( 2 x + 5 ) (5 - 2x)(5 + 2x) = -(2x - 5)(2x + 5) ( 5 − 2 x ) ( 5 + 2 x ) = − ( 2 x − 5 ) ( 2 x + 5 )
Cancel the common non-zero factor ( 2 x − 5 ) (2x - 5) ( 2 x − 5 ) : R ( x ) = ( 2 x − 5 ) ( x + 3 ) − ( 2 x − 5 ) ( 2 x + 5 ) = − x + 3 2 x + 5 R(x) = \frac{\cancel{(2x - 5)}(x + 3)}{-\cancel{(2x - 5)}(2x + 5)} = \mathbf{-\frac{x + 3}{2x + 5}} R ( x ) = − ( 2 x − 5 ) ( 2 x + 5 ) ( 2 x − 5 ) ( x + 3 ) = − 2x + 5 x + 3
State domain restrictions (where denominator was zero) : 25 − 4 x 2 ≠ 0 ⟹ x ≠ 5 2 , x ≠ − 5 2 25 - 4x^2 \neq 0 \implies x \neq \frac{5}{2}, \quad x \neq -\frac{5}{2} 25 − 4 x 2 = 0 ⟹ x = 2 5 , x = − 2 5
Common Pitfalls & ACCUPLACER Exam Traps
The "Freshman's Dream" Exponent Distribution Trap : Writing ( x − 4 ) 2 = x 2 − 16 (x - 4)^2 = x^2 - 16 ( x − 4 ) 2 = x 2 − 16 . Always write out binomial powers as multiplication: ( x − 4 ) ( x − 4 ) = x 2 − 8 x + 16 (x - 4)(x - 4) = x^2 - 8x + 16 ( x − 4 ) ( x − 4 ) = x 2 − 8 x + 16 .
Illegal Term-by-Term Cancellation : Canceling x 2 x^2 x 2 or numerical terms across addition in rational expressions (e.g., attempting to simplify x 2 + 5 x 2 + 2 \frac{x^2 + 5}{x^2 + 2} x 2 + 2 x 2 + 5 to 5 2 \frac{5}{2} 2 5 ). Terms cannot be canceled—only common factors.
Dropping the Negative Sign During GCF Extraction : In − 3 x 2 + 12 x − 15 -3x^2 + 12x - 15 − 3 x 2 + 12 x − 15 , factoring out − 3 -3 − 3 must flip all signs: − 3 ( x 2 − 4 x + 5 ) -3(x^2 - 4x + 5) − 3 ( x 2 − 4 x + 5 ) .
Neglecting to Factor Completely : Stopping at 2 x ( 4 x 2 − 36 ) 2x(4x^2 - 36) 2 x ( 4 x 2 − 36 ) without factoring out the common 4 to get 8 x ( x 2 − 9 ) = 8 x ( x − 3 ) ( x + 3 ) 8x(x^2 - 9) = 8x(x - 3)(x + 3) 8 x ( x 2 − 9 ) = 8 x ( x − 3 ) ( x + 3 ) .
Opposite Factor Sign Errors : Forgetting that x − y y − x = − 1 \frac{x - y}{y - x} = -1 y − x x − y = − 1 , not + 1 +1 + 1 .