4.2 Simplifying, Combining Like Terms, & Factoring

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.

$\text{Like Terms: } 7x^2 y^3 \quad \text{and} \quad -4x^2 y^3 \implies (7 - 4)x^2 y^3 = 3x^2 y^3$

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.

$\text{Unlike Terms: } 5x^2 \quad \text{and} \quad 5x \implies 5x^2 + 5x \quad (\text{Cannot be combined into } 10x^3 \text{ or } 10x^2)$

Comparison Table: Like vs. Unlike Terms

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The Distributive Property & Managing Signed Numbers

The Distributive Property allows a multiplier outside parentheses to be distributed to every term inside the parentheses:

$a(b + c) = ab + ac$ $a(b - c) = ab - ac$

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:

$-(ax^2 - bx + c) = -ax^2 + bx - c$ $-4(2x^2 - 5x + 3) = -4(2x^2) - 4(-5x) - 4(3) = -8x^2 + 20x - 12$

Simplifying Expressions with Nested Grouping Symbols

When simplifying expressions containing parentheses $( )$, brackets $[ ]$, or braces $\{ \}$, work systematically from the innermost grouping outward:

$\text{Expression: } 5 - 2[3x - 4(x - 2) + 7]$

1. Distribute innermost multiplier $-4$: $5 - 2[3x - 4x + 8 + 7]$
2. Combine like terms inside brackets: $5 - 2[-x + 15]$
3. Distribute outer multiplier $-2$: $5 + 2x - 30$
4. Combine constants: $2x - 25$

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Multiplying Polynomials

Multiplying polynomials requires applying the distributive property repeatedly along with the Product Rule of Exponents ($x^m \cdot x^n = x^{m+n}$).

1. Multiplying a Monomial by a Polynomial

Distribute the monomial to every term in the polynomial, multiplying numerical coefficients and adding variable exponents:

$-3x^2 y (4x^3 - 5xy^2 + 2y^3) = (-3x^2 y)(4x^3) + (-3x^2 y)(-5xy^2) + (-3x^2 y)(2y^3)$ $= -12x^5 y + 15x^3 y^3 - 6x^2 y^4$

2. Multiplying Two Binomials (The FOIL Method)

For two binomials $(a + b)(c + d)$, the FOIL acronym ensures all four pairwise products are calculated:

$(a + b)(c + d) = \underbrace{ac}_{\text{First}} + \underbrace{ad}_{\text{Outer}} + \underbrace{bc}_{\text{Inner}} + \underbrace{bd}_{\text{Last}}$

Worked Example: $(3x - 4)(2x + 5)$

• First: $(3x)(2x) = 6x^2$
• Outer: $(3x)(5) = +15x$
• Inner: $(-4)(2x) = -8x$
• Last: $(-4)(5) = -20$
• Combine middle like terms: $6x^2 + (15x - 8x) - 20 = \mathbf{6x^2 + 7x - 20}$

3. Special Polynomial Product Patterns

Mastering special product formulas allows instant expansion and forms the foundation for rapid factoring:

The Classic ACCUPLACER Trap: Exponents do never distribute across addition: $(a + b)^2 \neq a^2 + b^2$. Always include the middle cross-term $2ab$.

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:

$(2x - 3)(x^2 + 4x - 5) = 2x(x^2 + 4x - 5) - 3(x^2 + 4x - 5)$ $= (2x^3 + 8x^2 - 10x) - (3x^2 + 12x - 15)$ $= 2x^3 + (8x^2 - 3x^2) + (-10x - 12x) + 15$ $= \mathbf{2x^3 + 5x^2 - 22x + 15}$

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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).

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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: $\text{Factor: } 12x^4 y^3 - 18x^3 y^5 + 6x^2 y^2$

1. Numerical GCF of $12, 18, 6$ is $6$.
2. Lowest power of $x$ is $x^2$.
3. Lowest power of $y$ is $y^2$.
4. Overall $\text{GCF} = 6x^2 y^2$.
5. Divide each term by the GCF: $\frac{12x^4 y^3}{6x^2 y^2} = 2x^2 y, \quad \frac{-18x^3 y^5}{6x^2 y^2} = -3xy^3, \quad \frac{6x^2 y^2}{6x^2 y^2} = +1$
6. Factored form: $\mathbf{6x^2 y^2 (2x^2 y - 3xy^3 + 1)}$

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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: $\text{Factor: } 3x^3 - 6x^2 + 4x - 8$

1. Group into pairs: $(3x^3 - 6x^2) + (4x - 8)$
2. Factor GCF from first pair: $3x^2(x - 2)$
3. Factor GCF from second pair: $+4(x - 2)$
4. Factor out the common binomial $(x - 2)$: $\mathbf{(x - 2)(3x^2 + 4)}$

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Strategy 3: Factoring Quadratic Trinomials ($x^2 + bx + c$, Leading Coefficient $a = 1$)

To factor $x^2 + bx + c$, find two integers $p$ and $q$ that satisfy two simultaneous conditions:

1. Product: $p \cdot q = c$
2. Sum: $p + q = b$

Once found, the factored form is $(x + p)(x + q)$.

Sign Rules for Trinomial Factoring:

• If $c > 0$ and $b > 0$: Both $p$ and $q$ are positive ($+ , +$). Example: $x^2 + 8x + 12 = (x + 6)(x + 2)$
• If $c > 0$ and $b < 0$: Both $p$ and $q$ are negative ($- , -$). Example: $x^2 - 9x + 20 = (x - 4)(x - 5)$
• If $c < 0$: One factor is positive and one is negative ($+ , -$), with the larger absolute value matching the sign of $b$.
  • Example ($b > 0$): $x^2 + 2x - 15 = (x + 5)(x - 3)$
  • Example ($b < 0$): $x^2 - 5x - 24 = (x - 8)(x + 3)$

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Strategy 4: Factoring Quadratic Trinomials ($ax^2 + bx + c$, Leading Coefficient $a \neq 1$)

When $a \neq 1$ and cannot be factored out via GCF, use the $ac$-Method (Master Product Method):

1. Multiply $a$ and $c$ to obtain the target product $ac$.
2. Find two integers $p$ and $q$ whose product is $ac$ and whose sum is $b$.
3. Split the middle linear term $bx$ into $px + qx$.
4. Factor the resulting four-term polynomial by grouping.

Worked Example: $\text{Factor: } 6x^2 + 11x - 10$

1. Target product: $ac = 6 \cdot (-10) = -60$.
2. Target sum: $b = +11$.
3. Factor pairs of $-60$: Factors $+15$ and $-4$ satisfy $15 \cdot (-4) = -60$ and $15 + (-4) = 11$.
4. Rewrite middle term: $6x^2 + 15x - 4x - 10$
5. Factor by grouping: $3x(2x + 5) - 2(2x + 5)$
6. Extract common binomial: $\mathbf{(2x + 5)(3x - 2)}$

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Strategy 5: Factoring Difference of Two Squares ($a^2 - b^2$)

Any binomial consisting of two perfect squares separated by a minus sign factors into conjugate binomials:

$a^2 - b^2 = (a - b)(a + b)$

Worked Examples:

• $25x^2 - 49 = (5x)^2 - (7)^2 = \mathbf{(5x - 7)(5x + 7)}$
• Multi-Stage Factoring: $x^4 - 81 = (x^2 - 9)(x^2 + 9) = \mathbf{(x - 3)(x + 3)(x^2 + 9)}$

Important Note: The Sum of Squares ($a^2 + b^2$) is prime and cannot be factored over real numbers.

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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:

$\frac{P \cdot R}{Q \cdot R} = \frac{P}{Q} \quad (Q \neq 0, R \neq 0)$

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.

$\text{INCORRECT: } \frac{x^2 - 9}{x + 3} \neq x - 3 \text{ (by canceling } x^2/x \text{ and } -9/3)$ $\text{CORRECT: } \frac{x^2 - 9}{x + 3} = \frac{(x - 3)\cancel{(x + 3)}}{\cancel{x + 3}} = \mathbf{x - 3} \quad (x \neq -3)$

Managing Opposite Factors (The $-1$ Rule)

When a factor in the numerator is the exact negative of a factor in the denominator ($a - b$ versus $b - a$), factor out $-1$:

$b - a = -(a - b) \implies \frac{a - b}{b - a} = \frac{a - b}{-(a - b)} = \mathbf{-1}$

Worked Example: $\frac{2x - 6}{9 - x^2} = \frac{2(x - 3)}{(3 - x)(3 + x)} = \frac{2(x - 3)}{-(x - 3)(3 + x)} = \mathbf{-\frac{2}{x + 3}}$

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Comprehensive Step-by-Step Worked Examples

Worked Example 1: Multi-Step Expansion and Nested Simplification

Simplify the expression completely:

$E = 3(x - 4)^2 - 2(x + 2)(x - 5) - (x^2 - 14x + 8)$

Step-by-Step Solution:

1. Expand the squared binomial: $(x - 4)^2 = x^2 - 8x + 16$ $3(x - 4)^2 = 3(x^2 - 8x + 16) = 3x^2 - 24x + 48$

2. Expand the binomial product with FOIL: $(x + 2)(x - 5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10$ $-2(x + 2)(x - 5) = -2(x^2 - 3x - 10) = -2x^2 + 6x + 20$

3. Distribute the negative sign across the last polynomial: $-(x^2 - 14x + 8) = -x^2 + 14x - 8$

4. Combine all components: $E = (3x^2 - 24x + 48) + (-2x^2 + 6x + 20) + (-x^2 + 14x - 8)$

   • $x^2$ terms: $3x^2 - 2x^2 - x^2 = 0x^2 = 0$
   • $x$ terms: $-24x + 6x + 14x = -4x$
   • Constant terms: $48 + 20 - 8 = 60$

5. Final Result: $\mathbf{-4x + 60}$

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Worked Example 2: Complete Factoring of a Multi-Variable Expression

Factor the polynomial completely over the real numbers:

$P = 2x^4 y - 32y$

Step-by-Step Solution:

1. Factor out the GCF ($2y$): $P = 2y(x^4 - 16)$

2. Factor $x^4 - 16$ as a Difference of Two Squares: $x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4)$ $P = 2y(x^2 - 4)(x^2 + 4)$

3. Factor the remaining Difference of Squares ($x^2 - 4$): $x^2 - 4 = (x - 2)(x + 2)$ $P = 2y(x - 2)(x + 2)(x^2 + 4)$

4. Check the sum of squares factor: $(x^2 + 4)$ is prime over real numbers.

5. Final Factored Form: $\mathbf{2y(x - 2)(x + 2)(x^2 + 4)}$

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Worked Example 3: Simplifying a Rational Expression with $ac$-Factoring and Sign Inversion

Simplify the rational expression completely and state all domain restrictions:

$R(x) = \frac{2x^2 + x - 15}{25 - 4x^2}$

Step-by-Step Solution:

1. Factor the numerator $2x^2 + x - 15$ ($ac = 2(-15) = -30$, sum $= +1$):

   • Factors are $+6$ and $-5$: $2x^2 + 6x - 5x - 15 = 2x(x + 3) - 5(x + 3) = (2x - 5)(x + 3)$

2. Factor the denominator $25 - 4x^2$ (Difference of Squares): $25 - 4x^2 = 5^2 - (2x)^2 = (5 - 2x)(5 + 2x)$

3. Identify opposite binomial factors: Notice that $(5 - 2x) = -(2x - 5)$. Rewrite denominator: $(5 - 2x)(5 + 2x) = -(2x - 5)(2x + 5)$

4. Cancel the common non-zero factor $(2x - 5)$: $R(x) = \frac{\cancel{(2x - 5)}(x + 3)}{-\cancel{(2x - 5)}(2x + 5)} = \mathbf{-\frac{x + 3}{2x + 5}}$

5. State domain restrictions (where denominator was zero): $25 - 4x^2 \neq 0 \implies x \neq \frac{5}{2}, \quad x \neq -\frac{5}{2}$

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Common Pitfalls & ACCUPLACER Exam Traps

1. The "Freshman's Dream" Exponent Distribution Trap: Writing $(x - 4)^2 = x^2 - 16$. Always write out binomial powers as multiplication: $(x - 4)(x - 4) = x^2 - 8x + 16$.
2. Illegal Term-by-Term Cancellation: Canceling $x^2$ or numerical terms across addition in rational expressions (e.g., attempting to simplify $\frac{x^2 + 5}{x^2 + 2}$ to $\frac{5}{2}$). Terms cannot be canceled—only common factors.
3. Dropping the Negative Sign During GCF Extraction: In $-3x^2 + 12x - 15$, factoring out $-3$ must flip all signs: $-3(x^2 - 4x + 5)$.
4. Neglecting to Factor Completely: Stopping at $2x(4x^2 - 36)$ without factoring out the common 4 to get $8x(x^2 - 9) = 8x(x - 3)(x + 3)$.
5. Opposite Factor Sign Errors: Forgetting that $\frac{x - y}{y - x} = -1$, not $+1$.