2.4 Simplifying & Factoring Algebraic Expressions

2.4 Simplifying & Factoring Algebraic Expressions

Quick Summary: Simplifying and factoring algebraic expressions are fundamental techniques used to reduce complex mathematical models into manageable forms. On Part 1 of the EIAT, candidates are tested on expanding expressions using the FOIL method, combining like terms within nested grouping symbols, extracting the Greatest Common Factor (GCF), factoring quadratic trinomials ($ax^2 + bx + c$), and applying special patterns like the Difference of Squares ($a^2 - b^2$). In electrical theory, factoring enables rapid simplification of parallel resistance equations, impedance formulas, and power transfer expressions.

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Combining Like Terms & Handling Nested Grouping Symbols

An algebraic expression consists of numbers, variables, and operational signs connected without an equals sign (e.g., $3x^2 - 5xy + 4y^2$). Simplifying an expression means rewriting it in its most compact form by combining like terms and eliminating grouping symbols.

Defining Like Terms

Like terms are algebraic terms that contain the exact same variable bases raised to the exact same powers. Only the numerical coefficients differ.

• Like Terms: $4x^2y$ and $-7x^2y$ (same variables $x^2y$); $5ab$ and $12ab$.
• Unlike Terms: $3x^2$ and $3x$ (different powers of $x$); $4a^2b$ and $4ab^2$ (different variable exponents).

Rule: You can only add or subtract like terms! Unlike terms cannot be combined.

Handling Nested Grouping Symbols

When expressions contain nested grouping symbols such as parentheses $( )$, square brackets $[ ]$, and curly braces $\{ \}$, simplify from the innermost group working outward:

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

1. Distribute Innermost Parenthesis $-2(x - 5)$: $$2x - 3[ 4x - 2x + 10 ]$$
2. Combine Like Terms Inside Brackets $[ 4x - 2x + 10 ]$: $$2x - 3[ 2x + 10 ]$$
3. Distribute Outer $-3$ across Brackets: $$2x - 6x - 30$$
4. Combine Final Like Terms: $$-4x - 30$$

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Polynomial Expansion & The FOIL Method

Expanding algebraic expressions involves multiplying terms across parentheses using the distributive property.

Expanding Binomials: The FOIL Method

When multiplying two binomials $(a + b)(c + d)$, apply the FOIL acronym:

• F - First: Multiply the first terms in each binomial ($a \cdot c$).
• O - Outer: Multiply the outer terms ($a \cdot d$).
• I - Inner: Multiply the inner terms ($b \cdot c$).
• L - Last: Multiply the last terms in each binomial ($b \cdot d$).

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

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

1. First: $(2x) \cdot (3x) = 6x^2$
2. Outer: $(2x) \cdot (4) = 8x$
3. Inner: $(-5) \cdot (3x) = -15x$
4. Last: $(-5) \cdot (4) = -20$
5. Combine Middle Like Terms ($8x - 15x = -7x$): $$6x^2 - 7x - 20$$

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Factoring Techniques: Greatest Common Factor (GCF)

Factoring is the inverse process of polynomial expansion—it converts an expanded algebraic expression back into a product of simpler factors.

Extracting the Greatest Common Factor (GCF)

The first step in factoring ANY algebraic expression is to check for a Greatest Common Factor (GCF). The GCF is the largest numerical coefficient and variable power that divides evenly into every term of the expression.

Steps to Extract the GCF:

1. Find the GCF of Numerical Coefficients: Determine the greatest common divisor of all numbers.
2. Find the GCF of Variable Factors: Identify variables common to all terms and select the lowest exponent present for each variable.
3. Factor Out the GCF: Divide each original term by the GCF and write the expression as $\text{GCF} \times (\text{Remaining Terms})$.

Example: Factor $12x^3 y^2 - 18x^2 y^3 + 6x^2 y$

1. Numerical GCF of $12, 18, 6$: $6$.
2. Variable GCF of $x^3, x^2, x^2$: $x^2$ (lowest power).
3. Variable GCF of $y^2, y^3, y^1$: $y$ (lowest power).
4. Overall GCF: $6x^2y$.
5. Divide Terms by GCF: $$\frac{12x^3 y^2}{6x^2 y} = 2xy, \quad \frac{-18x^2 y^3}{6x^2 y} = -3y^2, \quad \frac{6x^2 y}{6x^2 y} = 1$$
6. Factored Result: $$6x^2 y (2xy - 3y^2 + 1)$$

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Factoring Quadratic Trinomials ($ax^2 + bx + c$)

A quadratic trinomial is a second-degree polynomial of the form $ax^2 + bx + c$.

Case 1: Simple Quadratic Trinomials ($a = 1$)

When the leading coefficient $a = 1$, the expression simplifies to $x^2 + bx + c$.

The Product-Sum Method:

To factor $x^2 + bx + c$, find two numbers $p$ and $q$ such that:

• Their product equals $c$ ($p \cdot q = c$).
• Their sum equals $b$ ($p + q = b$).

Once $p$ and $q$ are found, the factored form is:

$$(x + p)(x + q)$$

Example: Factor $x^2 - 9x + 20$

1. Identify $b = -9$ and $c = +20$.
2. List integer pairs that multiply to $+20$:
   • $(1, 20) \implies \text{Sum } 21$
   • $(2, 10) \implies \text{Sum } 12$
   • $(4, 5) \implies \text{Sum } 9$
   • $(-4, -5) \implies \text{Sum } -9$ (Match!)
3. Factored form: $$(x - 4)(x - 5)$$

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Case 2: General Quadratic Trinomials ($a \neq 1$) via the AC Method

When the leading coefficient $a \neq 1$, use the AC Method (Factoring by Grouping):

Steps for the AC Method ($ax^2 + bx + c$):

1. Calculate the Product $a \cdot c$.
2. Find Two Numbers that Multiply to $ac$ and Add to $b$.
3. Rewrite the Middle Term $bx$ using these two numbers.
4. Factor by Grouping (group the first two terms and the last two terms).

Example: Factor $2x^2 + 7x + 6$

1. Calculate $a \cdot c = 2 \cdot 6 = 12$.
2. Find factors of $12$ that add to $b = 7$: The numbers are $3$ and $4$ ($3 \cdot 4 = 12$ and $3 + 4 = 7$).
3. Rewrite middle term $7x$ as $3x + 4x$: $$2x^2 + 4x + 3x + 6$$
4. Group and factor: $$(2x^2 + 4x) + (3x + 6) = 2x(x + 2) + 3(x + 2)$$
5. Factor out common binomial $(x + 2)$: $$(x + 2)(2x + 3)$$

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Special Factoring Formulas

Recognizing special polynomial identities allows instantaneous factoring without trial and error.

1. Difference of Squares Formula

The difference of two perfect squares factors into the product of two conjugate binomials:

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

Warning: A Sum of Squares ($a^2 + b^2$) cannot be factored over real numbers!

Example: Factor $4x^2 - 25$

• $\sqrt{4x^2} = 2x$ and $\sqrt{25} = 5$.
• Apply formula: $(2x - 5)(2x + 5)$.

2. Perfect Square Trinomial Formulas

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

3. Multi-Step Factoring (GCF + Special Patterns)

Always check for a GCF first before applying special factoring formulas!

Example: Factor $2x^3 - 18x$ completely

1. Extract GCF of $2x$: $$2x^3 - 18x = 2x(x^2 - 9)$$
2. Recognize difference of squares in $(x^2 - 9)$: $$x^2 - 9 = (x - 3)(x + 3)$$
3. Combine for complete factored form: $$2x(x - 3)(x + 3)$$

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Electrical Trade Application: Parallel Resistance & Algebraic Formulas

In electrical engineering and circuit design, simplifying rational algebraic expressions is crucial when calculating equivalent resistance ($R_{eq}$) in parallel circuits.

Parallel Resistance Formula Reduction

For two resistors $R_1$ and $R_2$ connected in parallel, the reciprocal formula is:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$

To combine the right-hand side, find a common denominator ($R_1 R_2$):

$$\frac{1}{R_{eq}} = \frac{R_2}{R_1 R_2} + \frac{R_1}{R_1 R_2} = \frac{R_1 + R_2}{R_1 R_2}$$

Taking the reciprocal yields the famous Product-over-Sum Formula:

$$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$$

Substituting Power Formulas via Algebraic Factoring

Similarly, combining Ohm's Law ($V = IR$) with Joule's Law ($P = VI$) involves algebraic substitution:

$$P = V \cdot I = (I \cdot R) \cdot I = I^2 R$$ $$P = V \cdot I = V \cdot \left(\frac{V}{R}\right) = \frac{V^2}{R}$$

Factoring common variables from complex power expressions allows electricians to analyze voltage drops across multi-resistor branch networks efficiently.

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Step-by-Step Worked Examples

Worked Example 1: Multi-Term Nested Distribution

Problem: Simplify the algebraic expression completely:

$$3(2a^2 - 3ab + 4b^2) + 4(a^2 - ab + 2b^2)$$

Solution:

1. Distribute 3 across First Trinomial: $$3 \cdot 2a^2 - 3 \cdot 3ab + 3 \cdot 4b^2 = 6a^2 - 9ab + 12b^2$$
2. Distribute 4 across Second Trinomial: $$4 \cdot a^2 - 4 \cdot ab + 4 \cdot 2b^2 = 4a^2 - 4ab + 8b^2$$
3. Combine Expanded Expressions: $$(6a^2 - 9ab + 12b^2) + (4a^2 - 4ab + 8b^2)$$
4. Group Like Terms ($a^2$, $ab$, $b^2$): $$(6a^2 + 4a^2) + (-9ab - 4ab) + (12b^2 + 8b^2)$$
5. Add Coefficients: $$10a^2 - 13ab + 20b^2$$

Final Answer: $10a^2 - 13ab + 20b^2$

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Worked Example 2: Factoring a Quadratic Trinomial

Problem: Factor the quadratic trinomial completely:

$$x^2 - 9x + 20$$

Solution:

1. Identify Coefficients: $a = 1$, $b = -9$, $c = +20$.
2. Find Two Numbers $p$ and $q$ that Multiply to $+20$ and Add to $-9$: $$\text{Factors of } +20: (-1, -20), (-2, -10), (-4, -5)$$ $$\text{Test Sums: } (-1) + (-20) = -21, \quad (-2) + (-10) = -12, \quad (-4) + (-5) = -9$$
3. Select Numbers: $p = -4$ and $q = -5$.
4. Write Factored Form: $$(x - 4)(x - 5)$$

Final Answer: $(x - 4)(x - 5)$

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Worked Example 3: GCF Extraction Followed by Difference of Squares

Problem: Factor the polynomial expression completely:

$$2x^3 - 18x$$

Solution:

1. Identify GCF: Both terms $2x^3$ and $-18x$ share a numerical factor of $2$ and variable factor of $x$. GCF = $2x$.
2. Factor Out GCF: $$2x^3 - 18x = 2x(x^2 - 9)$$
3. Analyze Polynomial Inside Parentheses ($x^2 - 9$): Recognize $x^2 - 9$ as a Difference of Squares where $a = x$ and $b = 3$. $$x^2 - 9 = (x - 3)(x + 3)$$
4. Combine All Factors: $$2x(x - 3)(x + 3)$$

Final Answer: $2x(x - 3)(x + 3)$

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Worked Example 4: Algebraic Reduction of Equivalent Resistance

Problem: Two parallel branch resistors have values defined by algebraic expressions $R_1 = 2x$ and $R_2 = 4x$. Write a simplified single algebraic expression for the total equivalent parallel resistance $R_{eq}$.

Solution:

1. Use Product-over-Sum Parallel Resistance Formula: $$R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$$
2. Substitute $R_1 = 2x$ and $R_2 = 4x$: $$R_{eq} = \frac{(2x) \cdot (4x)}{2x + 4x}$$
3. Multiply Numerator and Add Denominator: $$R_{eq} = \frac{8x^2}{6x}$$
4. Simplify Coefficient Fraction and Cancel Common Variable $x$: $$R_{eq} = \left(\frac{8}{6}\right) \cdot \frac{x^2}{x} = \frac{4}{3} x = 1.33x$$

Final Answer: $\frac{4}{3} x$ (or $\frac{4x}{3}$)