Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Simplify expressions by combining like terms.
• Expand expressions with single and binomial factors.
• Factorise expressions using common factors and difference of squares.
• Recognize and use special products like (a+b)² and (a-b)².
• Apply these skills to solve problems.

Lesson summary

Algebraic expressions form the bedrock of higher-level mathematics. Mastering them is crucial for success in subjects like Physics, Chemistry, and Accounting. In everyday life, we use algebraic thinking to solve problems involving budgets, discounts, and proportions. This week, we'll revisit and solidify your understanding of algebraic expressions, focusing on simplifying, expanding, and factorising. We'll examine how these techniques can be applied to solve real-world problems relevant to the South African context. Understanding algebraic expressions allows you to model situations mathematically and find solutions efficiently.

Lesson notes

2.1 Basic Terminology Variable: A letter (e.g., x, y, a, b) representing an unknown value.

Constant: A number that does not change (e.g., 5, -3, ½).

Coefficient: The number multiplying a variable (e.g., in 5x, 5 is the coefficient).

Algebraic Expression: A combination of terms connected by mathematical operations (+, -, ×, ÷).

Examples: 2x + 3y - 5, a² - 4b.

Like Terms: Terms that have the same variable(s) raised to the same power (e.g., 3x and -7x are like terms, but 3x and 3x² are not). 2.2 Simplifying Algebraic Expressions Simplifying involves combining like terms. Remember to pay close attention to the signs (+ or -) in front of each term.

Example 1: Simplify 5x + 3y - 2x + 7y - x Step 1: Group like terms together: (5x - 2x - x) + (3y + 7y)

Step 2: Combine the coefficients of like terms: (5 - 2 - 1)x + (3 + 7)y Step 3: Simplify: 2x + 10y Example 2: Simplify 3a²b - 5ab² + 2a²b + ab² - a²b Step 1: Group like terms together: (3a²b + 2a²b - a²b) + (-5ab² + ab²)

Step 2: Combine coefficients: (3 + 2 - 1)a²b + (-5 + 1)ab² Step 3: Simplify: 4a²b - 4ab² 2.3 Expanding Algebraic Expressions Expanding involves removing brackets by multiplying each term inside the bracket by the term outside the bracket.

This uses the distributive property: a(b + c) = ab + ac Example 3: Expand 3(2x - 5)

Step 1: Multiply 3 by each term inside the bracket: 3 2x - 3 * 5 Step 2: Simplify: 6x - 15 Example 4: Expand -2a(a² - 3a + 4)

Step 1: Multiply -2a by each term inside the bracket: -2a a² + (-2a) (-3a) + (-2a) 4 Step 2: Simplify: -2a³ + 6a² - 8a 2.4 Expanding Binomial Products When multiplying two binomials (expressions with two terms each), we use the FOIL method (First, Outer, Inner, Last) or the distributive property twice.

Example 5: Expand (x + 2)(x - 3)

Method 1: FOIL First: x x = x² Outer: x -3 = -3x Inner: 2 x = 2x Last: 2 -3 = -6 Step 2: Combine the terms: x² - 3x + 2x - 6 Step 3: Simplify: x² - x - 6 Method 2: Distributive Property x(x-3) + 2(x-3) x^2 - 3x + 2x - 6 x^2 - x - 6 Example 6: Expand (2a - 1)(3a + 4)

FOIL: F: 2a 3a = 6a² O: 2a 4 = 8a I: -1 3a = -3a L: -1 4 = -4 Step 2: Combine terms: 6a² + 8a - 3a - 4 Step 3: Simplify: 6a² + 5a - 4 2.5 Special Products Certain binomial products occur frequently and have predictable patterns: (a + b)² = a² + 2ab + b² (a - b)² = a² - 2ab + b² (a + b)(a - b) = a² - b² (Difference of Squares)

Example 7: Expand (x + 3)² Using the formula: (x + 3)² = x² + 2(x)(3) + 3² Simplify: x² + 6x + 9 Example 8: Expand (2y - 5)² Using the formula: (2y - 5)² = (2y)² - 2(2y)(5) + 5² Simplify: 4y² - 20y + 25 Example 9: Expand (a + 4)(a - 4)

Using the formula: (a + 4)(a - 4) = a² - 4² Simplify: a² - 16 2.6 Factorising Algebraic Expressions Factorising is the reverse of expanding. It involves writing an expression as a product of factors. 2.6.1 Highest Common Factor (HCF) Find the HCF of all the terms in the expression. Divide each term by the HCF and write the HCF outside the bracket.

Example 10: Factorise 6x + 9 Step 1: Find the HCF of 6 and 9, which is

3. Step 2: Divide each term by 3: (6x / 3) + (9 / 3) = 2x + 3 Step 3: Write the expression as a product: 3(2x + 3)

Example 11: Factorise 12a²b - 8ab² + 4ab Step 1: Find the HCF of 12, 8, and 4, which is

4. The HCF of a²b, ab², and ab is ab.

Therefore, the HCF of the whole expression is 4ab.

Step 2: Divide each term by 4ab: (12a²b / 4ab) - (8ab² / 4ab) + (4ab / 4ab) = 3a - 2b + 1 Step 3: Write the expression as a product: 4ab(3a - 2b + 1) 2.6.2 Difference of Squares If the expression is in the form a² - b², it can be factorised as (a + b)(a - b).

Example 12: Factorise x² - 25 Step 1: Recognise that x² - 25 is in the form a² - b², where a = x and b =

5. Step 2: Apply the formula: (x + 5)(x - 5)

Example 13: Factorise 4y² - 9 Step 1: Recognise that 4y² - 9 is in the form a² - b², where a = 2y and b =

3. Step 2: Apply the formula: (2y + 3)(2y - 3) Guided Practice (With Solutions)

Question 1: Simplify: 7p - 4q + 2p + 6q - 3p Solution: Step 1: Group like terms: (7p + 2p - 3p) + (-4q + 6q)

Step 2: Combine coefficients: (7 + 2 - 3)p + (-4 + 6)q Step 3: Simplify: 6p + 2q

Commentary: This question reinforces the concept of identifying and combining like terms. Pay attention to the signs (+ or -) before each term when grouping them.* Question 2: Expand: -5x(2x² - x + 3)

Solution: Step 1: Multiply -5x by each term inside the bracket: -5x 2x² + (-5x) (-x) + (-5x) 3 Step 2: Simplify: -10x³ + 5x² - 15x

Commentary: This question tests your understanding of the distributive property and how to handle negative signs when multiplying.* Question 3: Factorise: 15a²b + 25ab² - 10ab Solution: Step 1: Find the HCF of 15, 25, and 10, which is

5. The HCF of a²b, ab², and ab is ab.

Therefore, the HCF of the whole expression is 5ab.