Algebraic Expressions and Identities — Class 8 Mathematics

1. Basic Terminology

Algebraic Expression: A combination of constants and variables connected by +, −, ×, ÷. Examples: 3x+5, 2a²−3b+7, xy+yz+zx.

Term: A single part of an expression separated by + or −. In 3x²+2xy−5, the terms are 3x², 2xy, −5.

Coefficient: The NUMERICAL FACTOR of a term. In 7xy²: coefficient = 7 (numerical), x (literal coefficient of y²), y² (literal coefficient of x).

Types by number of terms: Monomial — ONE term (4x, −7a²b). Binomial — TWO terms (x+5, a²−b²). Trinomial — THREE terms (x²+2x+1). Polynomial — MANY terms (general term for one or more terms).

2. Like and Unlike Terms

Like terms have the SAME variables raised to the SAME powers. 3x²y and −5x²y are LIKE. 3x²y and 3xy² are UNLIKE (powers differ). Only LIKE terms can be added or subtracted: 3x²+5x² = 8x². 3x²+5x cannot be combined further.

3. Operations on Algebraic Expressions

Addition

Add the COEFFICIENTS of like terms. Keep unlike terms unchanged. (3x²+2x−5) + (4x²−3x+7) = (3x²+4x²) + (2x−3x) + (−5+7) = 7x²−x+2.

Subtraction

Change the sign of EVERY term being subtracted. Then add. (5a−3b+2) − (2a+b−4) = 5a−3b+2−2a−b+4 = 3a−4b+6.

Multiplication

Multiply coefficients AND variables. For variables: x^m × x^n = x^(m+n). A monomial × a binomial uses DISTRIBUTIVE PROPERTY: 3x(2x+5) = 3x×2x + 3x×5 = 6x²+15x. A binomial × a binomial uses DOUBLE distributive (FOIL): (x+3)(x+5) = x(x+5)+3(x+5) = x²+5x+3x+15 = x²+8x+15.

4. Standard Identities

An IDENTITY is an equation TRUE for ALL values of the variable(s). Identities are TOOLS — they let you expand or factorise in one step.

Identity 1: (a+b)² = a² + 2ab + b²

Derivation: (a+b)(a+b) = a(a+b) + b(a+b) = a²+ab+ba+b² = a²+2ab+b². 'The square of a sum = square of first + twice product + square of second.'

Geometric meaning: Area of a square of side (a+b) = a² + ab + ab + b² = a² + 2ab + b².

Application: (x+5)² = x² + 2(x)(5) + 5² = x²+10x+25. 103² = (100+3)² = 10000+600+9 = 10609.

Identity 2: (a−b)² = a² − 2ab + b²

Derivation: (a−b)(a−b) = a²−ab−ba+b² = a²−2ab+b². 'The negative sign only affects the middle term — the a² and b² remain positive.'

Application: (x−7)² = x²−14x+49. 98² = (100−2)² = 10000−400+4 = 9604.

Identity 3: (a+b)(a−b) = a² − b²

Derivation: (a+b)(a−b) = a²−ab+ba−b² = a²−b². 'The middle terms CANCEL. This is the DIFFERENCE OF SQUARES — the most useful identity.'

Application: (x+8)(x−8) = x²−64. 105×95 = (100+5)(100−5) = 10000−25 = 9975.

Identity 4: (x+a)(x+b) = x² + (a+b)x + ab

Derivation: x² + xb + ax + ab = x² + (a+b)x + ab. 'The coefficient of x is the SUM of a and b. The constant term is the PRODUCT of a and b.'

Application: (x+3)(x+7) = x² + 10x + 21. (y−4)(y+6) = y² + 2y − 24.

5. Applying Identities — Worked Examples

Example 1 — Expand using Identity: (3x+4y)². Identity 1: a=3x, b=4y. (3x)² + 2(3x)(4y) + (4y)² = 9x² + 24xy + 16y².

Example 2 — Factorise using Identity: 25x²−49y². Identity 3 (difference of squares): a²−b² = (a+b)(a−b). a=5x, b=7y. 25x²−49y² = (5x+7y)(5x−7y).

Example 3 — Evaluate: (7.5)²−(2.5)². Identity 3: = (7.5+2.5)(7.5−2.5) = 10×5 = 50.

Example 4 — Missing term: Fill the blank: (___ + 3)² = x² + 12x + 9. Comparing with Identity 1: a² = x² → a = x. 2ab = 12x → 2(x)(3) = 6x... wait. Actually: 2ab = 12x, and b=3, so 2×a×3 = 6a → a = 2x. No: (2x+3)² = 4x²+12x+9, not x²+12x+9. Hmm: Comparing directly: a² = x² so a = x. 2ab = 12x → 2(x)(b) = 12x → b = 6. b² = 36, but RHS shows 9. This doesn't work — the RHS must be checked.

6. Common Mistakes

1. (a+b)² ≠ a²+b²: (3+4)² = 49, but 3²+4² = 25. The 2ab term is MISSING.
2. (a−b)² ≠ a²−b²: (5−2)² = 9, but 5²−2² = 21. Again, the −2ab term is needed.
3. Sign error in Identity 2: (x−5)² = x²−10x+25. The LAST term is POSITIVE (because (−5)²=+25).
4. Forgetting to multiply ALL terms when expanding: 2x(x+3) = 2x²+6x. Not 2x²+3 or 2x+6x.

7. AP Exam Focus

Key Exam Tips

• For 'evaluate using identities': identify WHICH identity. Write it. Substitute. Simplify. Show ALL steps.
• (x+a)(x+b): sum = a+b (coefficient of x). Product = ab (constant term). Use this for quick mental expansion.
• The difference of squares identity (a+b)(a−b) = a²−b² is the MOST versatile — it works for ANY conjugate pair.

Quick Self-Test

1. Coefficient of x in 3x²−4x+7? (Answer: −4.)
2. (2x+3y)² = ? (Answer: 4x²+12xy+9y².)
3. (p−6)(p+6) = ? (Answer: p²−36.)
4. Use identity to evaluate 97×103. (Answer: (100−3)(100+3)=10000−9=9991.)
5. Factorise: 9x²−16y². (Answer: (3x+4y)(3x−4y).)
6. Expand (y+2)(y+5). (Answer: y²+7y+10.)

More Applications of Identities

Evaluating Squares Mentally: 205² = (200+5)² = 40000+2000+25 = 42025. 198² = (200−2)² = 40000−800+4 = 39204. 47² = (50−3)² = 2500−300+9 = 2209. 'The identities transform a hard square calculation into easy arithmetic. This is how mental calculators do it.'

Evaluating Products: 101×99 = (100+1)(100−1) = 10000−1 = 9999. 53×47 = (50+3)(50−3) = 2500−9 = 2491. 35×45 = (40−5)(40+5) = 1600−25 = 1575. 'Any two numbers at equal distance from a central number multiply as difference of squares.'

Using (x+a)(x+b) for Quick Multiplication: 107×103 = (100+7)(100+3) = 100²+(7+3)×100+7×3 = 10000+1000+21 = 11021. 34×36 = (30+4)(30+6) = 900+(4+6)×30+24 = 900+300+24 = 1224.

Geometric Verification of Identities

• (a+b)²: Draw a square of side (a+b). It contains: a square a², a square b², and two rectangles of area ab each. Total = a²+2ab+b².
• (a−b)²: From a square a², remove two rectangles ab and add back b² (which was subtracted twice). Result = a²−2ab+b².
• (a+b)(a−b): From a square a², remove b². The remaining L-shaped figure has area (a+b)(a−b). 'Geometric proofs make identities VISUAL. They're not just algebra — they're FACTS about AREAS.'

Simplifying Complex Expressions

Simplify: (x+2)² + (x−2)². = (x²+4x+4) + (x²−4x+4) = 2x²+8 = 2(x²+4). 'Adding the squares of a binomial and its conjugate ELIMINATES the middle terms.'

Simplify: (x+3)² − (x−3)². = (x²+6x+9)−(x²−6x+9) = 12x. 'The difference of squares of conjugates simplifies to 4ab: (a+b)²−(a−b)² = 4ab.'

Common Errors Deep Dive

A student writes (x+5)² = x²+25. The 2×x×5 = 10x term is MISSING. 'Memorise the full identity. (a+b)² = a²+2ab+b². The middle term 2ab is NOT optional.'

A student factorises x²−4 as (x−2)². Expand to check: (x−2)² = x²−4x+4 ≠ x²−4. 'a²−b² = (a+b)(a−b), not (a−b)². These two identities are DIFFERENT. Verify by expanding.'

Key Exam Tips

• When asked 'Evaluate using suitable identity,' explicitly STATE which identity you are using.
• For 'simplify' questions: show each step, combining like terms.
• Always present the final answer in DESCENDING powers of the variable: x² term, then x term, then constant.
• Check: if x=1 gives LHS=RHS, your simplification is probably correct. This is a quick numerical verification.