Expression Work. Grade 7 Mathematics: Expression Work - Expanding and Factorising Subtopic Navigator Introduction to Expression Work Expanding Single Brackets The Distributive Law Expanding Multiple Brackets Introduction to Factorising Factorising by Common Factors Basic Factorising Techniques Checking Your Work Applications and Problem Solving Practice Exercises Conclusion Lesson Objectives Understand and apply the distributive law to expand expressions Expand single brackets with both positive and negative terms Expand expressions with multiple brackets Understand the concept of factorising as reverse expanding Identify common factors in algebraic expressions Factorise simple expressions by taking out common factors Check work by expanding factorised expressions Apply expanding and factorising to solve problems Introduction to Expression Work Expanding and factorising are two fundamental skills in algebra that work in opposite directions. Expanding means removing brackets by multiplying out terms. Factorising means putting brackets back in by finding common factors. These skills are essential for simplifying expressions, solving equations, and working with formulas. Key Relationship: Expanding and factorising are inverse operations: If $3(x + 4) = 3x + 12$ (expanding) Then $3x + 12 = 3(x + 4)$ (factorising) Expanding Single Brackets Expanding a single bracket means multiplying everything inside the bracket by the term outside. We use the distributive law: $a(b + c) = ab + ac$ Example 1: Basic Expanding Expand: $3(x + 5)$ Solution: Multiply 3 by each term inside the bracket: $3 times x = 3x$ $3 times 5 = 15$ So: $3(x + 5) = 3x + 15$ Example 2: With Negative Terms Expand: $2(3x - 4)$ Solution: Multiply 2 by each term: $2 times 3x = 6x$ $2 times (-4) = -8$ (careful with the negative!) So: $2(3x - 4) = 6x - 8$ Expanding Practice Expand: $4(x + 2)$ Expand: $5(2y - 3)$ Expand: $2(3a + 4b)$ Expand: $3(4x - 5)$ Expand: $7(2m + 1)$ The Distributive Law The distributive law states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products. This works for both positive and negative coefficients. Distributive Law Formulas: $a(b + c) = ab + ac$ $a(b - c) = ab - ac$ $(b + c)a = ba + ca$ $(b - c)a = ba - ca$ The order doesn't matter: $a(b + c) = (b + c)a$ Example 1: Negative Outside Bracket Expand: $-3(2x - 4)$ Solution: Multiply -3 by each term: $-3 times 2x = -6x$ $-3 times (-4) = +12$ (negative × negative = positive) So: $-3(2x - 4) = -6x + 12$ Example 2: Variable Outside Bracket Expand: $x(3x + 2)$ Solution: Multiply $x$ by each term: $x times 3x = 3x^2$ $x times 2 = 2x$ So: $x(3x + 2) = 3x^2 + 2x$ Common Mistake: When expanding $2(3x + 4)$, students often write $6x + 4$ (forgetting to multiply the 4 by 2). Correct: $2(3x + 4) = 6x + 8$ Remember: Multiply EVERY term inside the bracket! Distributive Law Practice Expand: $-2(x + 3)$ Expand: $-4(2y - 5)$ Expand: $x(x + 4)$ Expand: $2x(3x - 1)$ Expand: $-3(4 - 2x)$ Expanding Multiple Brackets When there are multiple terms outside a bracket, or brackets inside brackets, we expand step by step, following the order of operations. Example 1: Multiple Terms Outside Expand: $2x(3x + 4) - 3(2x - 1)$ Solution: Expand each part separately: $2x(3x + 4) = 6x^2 + 8x$ $-3(2x - 1) = -6x + 3$ Combine: $6x^2 + 8x - 6x + 3$ Simplify: $6x^2 + 2x + 3$ Example 2: Nested Brackets Expand: $2(3(x + 2) - 4)$ Solution: Work from inside out: First: $3(x + 2) = 3x + 6$ Then: $3x + 6 - 4 = 3x + 2$ Finally: $2(3x + 2) = 6x + 4$ Or expand step by step: $2(3(x + 2) - 4) = 2(3x + 6 - 4) = 2(3x + 2) = 6x + 4$ Multiple Brackets Practice Expand: $3(2x + 1) + 4(x - 2)$ Expand: $2x(x + 3) - x(2x - 1)$ Expand: $4(2(x - 1) + 3)$ Expand: $3(2y + 4) - 2(3y - 5)$ Expand: $x(2x + 3) + 2(3x - 4)$ Introduction to Factorising Factorising is the reverse of expanding. Instead of removing brackets, we put brackets in by finding common factors. A factorised expression is usually simpler and more useful. What is Factorising? Factorising means writing an expression as a product of its factors. Example: $6x + 12 = 6(x + 2)$ Here, 6 is the common factor, and $(x + 2)$ is the other factor. Example 1: Simple Factorising Factorise: $4x + 8$ Solution: Look for common factors of both terms: $4x = 4 times x$ $8 = 4 times 2$ Both have factor 4 So: $4x + 8 = 4(x + 2)$ Check by expanding: $4(x + 2) = 4x + 8$ ✓ Example 2: Factorising with Variables Factorise: $3x^2 + 6x$ Solution: Common factors: Both have $3$ and $x$ $3x^2 = 3x times x$ $6x = 3x times 2$ So: $3x^2 + 6x = 3x(x + 2)$ Check: $3x(x + 2) = 3x^2 + 6x$ ✓ Factorising Practice Factorise: $5x + 10$ Factorise: $6y - 12$ Factorise: $4x^2 + 8x$ Factorise: $2a^2 - 4a$ Factorise: $7m + 14$ Factorising by Common Factors To factorise by taking out common factors: 1. Find the highest common factor (HCF) of all terms 2. Write the HCF outside the bracket 3. Divide each term by the HCF to find what goes inside Step-by-Step Factorising: 1. Identify common factors in all terms 2. Find the Highest Common Factor (HCF) 3. Write HCF outside brackets 4. Divide each term by HCF to get inside brackets 5. Check by expanding back Example 1: Finding HCF Factorise: $12x + 18$ Solution: Find HCF of 12 and 18: Factors of 12: 1, 2, 3, 4, 6, 12 Factors of 18: 1, 2, 3, 6, 9, 18 HCF = 6 Divide each term by 6: $12x ÷ 6 = 2x$ $18 ÷ 6 = 3$ So: $12x + 18 = 6(2x + 3)$ Example 2: With Multiple Variables Factorise: $8x^2y + 12xy^2$ Solution: Find common factors: Numerical: HCF of 8 and 12 is 4 Variables: Both have $x$ and $y$ Common factor = $4xy$ Divide each term by $4xy$: $8x^2y ÷ 4xy = 2x$ $12xy^2 ÷ 4xy = 3y$ So: $8x^2y + 12xy^2 = 4xy(2x + 3y)$ Common Factor Practice Factorise: $15x + 25$ Factorise: $18y - 24$ Factorise: $9x^2 + 15x$ Factorise: $12a^2b - 18ab^2$ Factorise: $20mn + 25n$ Basic Factorising Techniques Basic factorising involves recognizing patterns and taking out common factors. Always check if there's a common factor in all terms first. Example 1: Factorising with Negative Terms Factorise: $6x - 9$ Solution: HCF of 6 and 9 is 3 $6x ÷ 3 = 2x$ $-9 ÷ 3 = -3$ So: $6x - 9 = 3(2x - 3)$ Note: The negative sign goes inside the bracket. Example 2: When Common Factor is Negative Factorise: $-4x - 8$ Sometimes we factor out a negative for convenience. Solution: Option 1: Factor out 4: $4(-x - 2)$ Option 2: Factor out -4: $-4(x + 2)$ (often preferred) Both are correct, but $-4(x + 2)$ is usually simpler. Example 3: Three Terms Factorise: $3x^2 + 6x + 9$ Solution: Common factor in all three terms: $3x^2 = 3 times x^2$ $6x = 3 times 2x$ $9 = 3 times 3$ HCF = 3 Divide each term by 3: $3x^2 ÷ 3 = x^2$ $6x ÷ 3 = 2x$ $9 ÷ 3 = 3$ So: $3x^2 + 6x + 9 = 3(x^2 + 2x + 3)$ Basic Techniques Practice Factorise: $8x - 12$ Factorise: $-6y + 18$ Factorise: $5x^2 + 10x + 15$ Factorise: $2a^2 - 4a + 6$ Factorise: $-9m - 27$ Checking Your Work Always check your factorising by expanding back. If you get the original expression, your factorising is correct. This is an essential step to avoid errors. The Check-and-Balance Method: 1. Factorise the expression 2. Expand your answer 3. Compare with original expression 4. If they match, you're correct! Example 1: Checking Factorising Check if $4(2x + 3)$ is the correct factorisation of $8x + 12$ Solution: Expand $4(2x + 3)$: $4 times 2x = 8x$ $4 times 3 = 12$ $4(2x + 3) = 8x + 12$ ✓ This matches the original, so the factorisation is correct. Example 2: Finding an Error Someone factorised $6x + 9$ as $3(2x + 6)$. Check if this is correct. Solution: Expand $3(2x + 6)$: $3 times 2x = 6x$ $3 times 6 = 18$ $3(2x + 6) = 6x + 18$ This gives $6x + 18$, but original was $6x + 9$. Error: Should be $3(2x + 3)$ instead of $3(2x + 6)$ Common Checking Errors: 1. Forgetting to multiply all terms inside brackets 2. Making sign errors with negatives 3. Not finding the highest common factor 4. Missing variables in common factors Checking Practice Check: Is $5(2x + 1)$ the factorisation of $10x + 5$? Check: Is $3(3x + 4)$ the factorisation of $9x + 12$? Check: Is $2x(3x + 2)$ the factorisation of $6x^2 + 4x$? Check: Is $4(2x - 1)$ the factorisation of $8x - 4$? Check: Is $x(x + 3)$ the factorisation of $x^2 + 3x$? Applications and Problem Solving Expanding and factorising are not just abstract skills - they have practical applications in geometry, problem-solving, and simplifying calculations. Example 1: Geometry Application A rectangle has length $(x + 3)$ and width $(x + 2)$. Find and expand the expression for area. Solution: Area = length × width = $(x + 3)(x + 2)$ Expand using distributive law (double distribution): $(x + 3)(x + 2) = x(x + 2) + 3(x + 2)$ $= x^2 + 2x + 3x + 6$ $= x^2 + 5x + 6$ Area = $x^2 + 5x + 6$ square units Example 2: Problem Solving The perimeter of a square is $(12x + 16)$ cm. Find an expression for the side length. Solution: Perimeter of square = 4 × side length So side length = perimeter ÷ 4 Side length = $(12x + 16) ÷ 4$ Factorise: $12x + 16 = 4(3x + 4)$ So side length = $4(3x + 4) ÷ 4 = 3x + 4$ Each side = $(3x + 4)$ cm Application Practice Rectangle: length $(2x + 5)$, width $(x + 3)$. Find area expression and expand it. Triangle area = $frac{1}{2}bh$. If base = $(4x + 8)$ and height = $x$, write area expression and factorise. A number is $x$. Next number is $x + 1$. Their sum is $2x + 1$. Factorise this. Total cost of $x$ items at [latex]N5[/latex] each plus [latex]N3[/latex] delivery: $5x + 3$. Can this be factorised? Perimeter of rectangle: $2(3x + 4) + 2(2x - 1)$. Expand and simplify. Cumulative Exercises Expand: $3(2x + 5)$ Expand: $-2(4x - 3)$ Expand: $x(3x - 4)$ Expand: $2(3(x + 2) - 4)$ Expand: $3x(2x + 1) - 2(3x - 4)$ Factorise: $6x + 9$ Factorise: $8x^2 - 12x$ Factorise: $15y + 25$ Factorise: $4a^2b + 6ab^2$ Factorise: $9x^2 + 12x + 15$ Check by expanding: Is $5(3x + 2)$ equal to $15x + 10$? Rectangle area: $(x + 4)(x + 2)$. Expand this expression. The expression $12x + 18$ represents the total of 6 equal items. What is each item's value? Simplify by expanding: $2(3x + 1) + 3(2x - 4)$ Simplify by factorising first: $4x + 8 + 6x + 12$ Show/Hide Answers Problem: Expand: $3(2x + 5)$ Answer: $3 times 2x + 3 times 5 = 6x + 15$ Problem: Expand: $-2(4x - 3)$ Answer: $-2 times 4x + (-2) times (-3) = -8x + 6$ Problem: Expand: $x(3x - 4)$ Answer: $x times 3x + x times (-4) = 3x^2 - 4x$ Problem: Expand: $2(3(x + 2) - 4)$ Answer: First: $3(x + 2) = 3x + 6$; Then: $3x + 6 - 4 = 3x + 2$; Finally: $2(3x + 2) = 6x + 4$ Problem: Expand: $3x(2x + 1) - 2(3x - 4)$ Answer: $6x^2 + 3x - 6x + 8 = 6x^2 - 3x + 8$ Problem: Factorise: $6x + 9$ Answer: HCF = 3; $6x + 9 = 3(2x + 3)$ Problem: Factorise: $8x^2 - 12x$ Answer: Common factor = $4x$; $8x^2 - 12x = 4x(2x - 3)$ Problem: Factorise: $15y + 25$ Answer: HCF = 5; $15y + 25 = 5(3y + 5)$ Problem: Factorise: $4a^2b + 6ab^2$ Answer: Common factor = $2ab$; $4a^2b + 6ab^2 = 2ab(2a + 3b)$ Problem: Factorise: $9x^2 + 12x + 15$ Answer: HCF = 3; $9x^2 + 12x + 15 = 3(3x^2 + 4x + 5)$ Problem: Check by expanding: Is $5(3x + 2)$ equal to $15x + 10$? Answer: Expand: $5(3x + 2) = 15x + 10$. Yes, they are equal. Problem: Rectangle area: $(x + 4)(x + 2)$. Expand this expression. Answer: $(x + 4)(x + 2) = x(x + 2) + 4(x + 2) = x^2 + 2x + 4x + 8 = x^2 + 6x + 8$ Problem: The expression $12x + 18$ represents the total of 6 equal items. What is each item's value? Answer: Factorise: $12x + 18 = 6(2x + 3)$. Each item = $2x + 3$ Problem: Simplify by expanding: $2(3x + 1) + 3(2x - 4)$ Answer: $6x + 2 + 6x - 12 = 12x - 10$ Problem: Simplify by factorising first: $4x + 8 + 6x + 12$ Answer: Combine like terms: $10x + 20$; Factorise: $10(x + 2)$ Conclusion/Recap Expanding and factorising are essential algebraic skills that work in opposite directions. Expanding removes brackets by multiplication, while factorising puts brackets in by finding common factors. Key points to remember: 1. Expanding: Multiply each term inside brackets by the term outside 2. Distributive Law: $a(b + c) = ab + ac$ and $a(b - c) = ab - ac$ 3. Factorising: Find the highest common factor of all terms 4. Checking: Always expand your factorised expression to verify it matches the original 5. Applications: These skills are used in geometry, problem-solving, and simplifying expressions Mastering expanding and factorising provides a strong foundation for solving equations, working with formulas, and progressing to more advanced algebra in later grades. Clip It! Share your ANSWER in the Chat. Indicate TITLE e.g Linear Equation 1. .....2. e.t.c
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