Algebra

Polynomials, Equations, Exponents and Algebraic Identities

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Overview

Algebra forms the backbone of mathematics at the upper primary level (Classes 6–8) and is a high-scoring area in TN TET Paper II. This topic tests your ability to manipulate symbols, simplify expressions, solve equations and apply standard identities—skills every mathematics teacher must demonstrate confidently.

For TN TET, expect questions on identifying polynomial types, finding values of expressions using identities, solving linear equations and simplifying exponential expressions. The pedagogy section may also ask how to introduce algebraic concepts to young learners. Mastery here requires fluency with rules and the ability to spot shortcuts using identities.

The scope covers four interconnected areas: polynomials (classification and operations), equations (linear equations in one and two variables), exponents (laws of indices) and algebraic identities (standard expansions and factorisations). A strong grip on these fundamentals makes higher mathematics accessible to students.

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Key Concepts

• **Variable and Constant**: A variable (x, y) can take different values; a constant (3, –7) has a fixed value. Algebraic expressions combine both using operations.

• **Polynomial**: An expression with one or more terms where variables have whole-number exponents. Example: 3x² + 5x – 2. Polynomials are classified by degree (highest power) and number of terms.

• **Degree of a Polynomial**: The highest exponent of the variable. For 4x³ – x + 7, degree = 3. A constant non-zero polynomial has degree 0; zero polynomial has no defined degree.

• **Types by Terms**: Monomial (1 term), Binomial (2 terms), Trinomial (3 terms). Example: 5x is monomial; x + 1 is binomial; x² + x + 1 is trinomial.

• **Linear Equation**: Equation of degree 1. Standard form: ax + b = 0 (one variable) or ax + by + c = 0 (two variables). Solution is the value making LHS = RHS.

• **Exponent (Index/Power)**: In aⁿ, 'a' is base and 'n' is exponent. Exponents indicate repeated multiplication: 2⁴ = 2 × 2 × 2 × 2 = 16.

• **Algebraic Identity**: An equation true for all values of the variables. Unlike an equation (true for specific values), identities are universally valid and used for quick expansion/factorisation.

• **Zero of a Polynomial**: Value of variable that makes polynomial equal to zero. For p(x) = x – 3, zero is x = 3 because p(3) = 0.

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Formulas / Key Facts

### Laws of Exponents (a, b ≠ 0; m, n are integers)

| Law | Formula | |-----|---------| | Product Rule | aᵐ × aⁿ = aᵐ⁺ⁿ | | Quotient Rule | aᵐ ÷ aⁿ = aᵐ⁻ⁿ | | Power of Power | (aᵐ)ⁿ = aᵐⁿ | | Power of Product | (ab)ⁿ = aⁿbⁿ | | Power of Quotient | (a/b)ⁿ = aⁿ/bⁿ | | Zero Exponent | a⁰ = 1 | | Negative Exponent | a⁻ⁿ = 1/aⁿ |

### Standard Algebraic Identities

1. (a + b)² = a² + 2ab + b² 2. (a – b)² = a² – 2ab + b² 3. (a + b)(a – b) = a² – b² 4. (a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b) 5. (a – b)³ = a³ – 3a²b + 3ab² – b³ = a³ – b³ – 3ab(a – b) 6. a³ + b³ = (a + b)(a² – ab + b²) 7. a³ – b³ = (a – b)(a² + ab + b²) 8. (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

### Linear Equations

• One variable: ax + b = 0 → x = –b/a
• Two variables: Solved by substitution or elimination; graphically represented as straight lines.

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Worked Examples

### Example 1: Simplify using exponent laws **Simplify**: (2³ × 2⁵) ÷ 2⁴

**Solution**:

• Step 1: Apply product rule to numerator: 2³ × 2⁵ = 2³⁺⁵ = 2⁸
• Step 2: Apply quotient rule: 2⁸ ÷ 2⁴ = 2⁸⁻⁴ = 2⁴
• Step 3: Evaluate: 2⁴ = 16

**Answer**: 16

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### Example 2: Expand using identity **Expand**: (3x + 5)²

**Solution**:

• Use identity (a + b)² = a² + 2ab + b²
• Here a = 3x, b = 5
• (3x)² + 2(3x)(5) + (5)²
• = 9x² + 30x + 25

**Answer**: 9x² + 30x + 25

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### Example 3: Factorise using identity **Factorise**: 49y² – 64

**Solution**:

• Recognise as difference of squares: a² – b²
• 49y² = (7y)² and 64 = 8²
• Apply identity: a² – b² = (a + b)(a – b)
• = (7y + 8)(7y – 8)

**Answer**: (7y + 8)(7y – 8)

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### Example 4: Solve linear equation **Solve**: 5x – 3 = 2x + 9

**Solution**:

• Step 1: Bring variable terms to one side: 5x – 2x = 9 + 3
• Step 2: Simplify: 3x = 12
• Step 3: Divide: x = 12/3 = 4

**Answer**: x = 4

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Common Mistakes

| Wrong Thinking | Correct Approach | |----------------|------------------| | Adding exponents when bases are different: 2³ × 3² = 6⁵ | Exponent rules apply only when bases are same. Evaluate separately: 8 × 9 = 72 | | Squaring a binomial as (a + b)² = a² + b² | Always include the middle term: (a + b)² = a² + **2ab** + b² | | Writing a⁰ = 0 | Any non-zero base raised to power 0 equals 1: a⁰ = 1 | | Confusing (–3)² with –3² | (–3)² = 9 (square of –3), but –3² = –9 (negative of 3²). Brackets matter! | | Applying a² – b² identity to a² + b² | Sum of squares does not factorise over real numbers. a² + b² ≠ (a + b)(a – b) |

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Quick Reference

• **Degree** = highest power of variable in a polynomial.
• **aᵐ × aⁿ = aᵐ⁺ⁿ** — add exponents when multiplying same bases.
• **(a + b)² = a² + 2ab + b²** — never forget the middle term.
• **(a + b)(a – b) = a² – b²** — instant factorisation for difference of squares.
• **a⁰ = 1** (a ≠ 0); **a⁻ⁿ = 1/aⁿ**.
• Linear equation solution: isolate variable by inverse operations.