Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Expand and simplify expressions with binomials and trinomials.
• Factorise using the difference of two squares (DOTS).
• Factorise by grouping in pairs.
• Solve problems using these algebraic skills.

Lesson summary

Algebraic expressions and factorization are fundamental building blocks in mathematics. They allow us to represent relationships between quantities and solve problems in a structured and efficient way. In South Africa, understanding these concepts is crucial for budgeting household income (algebraic expressions) and efficiently planning for land use in farming (factorisation, optimising area calculations), for instance. This week, we will delve deeper into more complex algebraic expressions and advanced factorization techniques, building on the knowledge you gained in previous weeks.

Lesson notes

2.1 Expanding and Simplifying Algebraic Expressions Expanding algebraic expressions involves removing brackets by multiplying each term inside the bracket by the term outside. Simplifying involves combining like terms (terms with the same variable and exponent). This builds on the distributive property we learned previously.

Example 1: Expanding binomials Consider the expression: (x + 2)(x + 3)

Step 1 (Distributive Property): Multiply the first term of the first bracket (x) by each term in the second bracket: x(x + 3) = x² + 3x Step 2 (Distributive Property): Multiply the second term of the first bracket (2) by each term in the second bracket: 2(x + 3) = 2x + 6 Step 3 (Combine): Add the results from steps 1 and 2: (x² + 3x) + (2x + 6)

Step 4 (Simplify): Combine like terms: x² + (3x + 2x) + 6 = x² + 5x + 6 Therefore, (x + 2)(x + 3) = x² + 5x + 6 Example 2: Expanding with negative signs: Consider the expression: (2a - 1)(a + 4)

Step 1: 2a(a + 4) = 2a² + 8a Step 2: -1(a + 4) = -a - 4 Step 3: (2a² + 8a) + (-a - 4)

Step 4: 2a² + 8a - a - 4 = 2a² + 7a - 4 Therefore, (2a - 1)(a + 4) = 2a² + 7a - 4 Example 3: Expanding binomial with trinomials: Consider the expression (x + 1)(x² + 2x - 3)

Step 1: x(x² + 2x - 3) = x³ + 2x² - 3x Step 2: 1(x² + 2x - 3) = x² + 2x - 3 Step 3: (x³ + 2x² - 3x) + (x² + 2x - 3)

Step 4: x³ + 2x² + x² - 3x + 2x - 3 = x³ + 3x² - x - 3 Therefore, (x + 1)(x² + 2x - 3) = x³ + 3x² - x - 3 2.2 Factorising using the Difference of Two Squares (DOTS) The difference of two squares (DOTS) is a special case of factorization. An expression in the form a² - b² can be factored as (a + b)(a - b).

Example 1: Factorise x² - 9 Step 1: Recognise that x² is a perfect square and 9 (which is 3²) is also a perfect square.

Step 2: Identify 'a' and 'b': a = x, b = 3 Step 3: Apply the formula: a² - b² = (a + b)(a - b)

Step 4: Substitute: x² - 9 = (x + 3)(x - 3)

Example 2: Factorise 4y² - 25 Step 1: 4y² is a perfect square ((2y)²) and 25 is a perfect square (5²).

Step 2: a = 2y, b = 5 Step 3: (a + b)(a - b)

Step 4: 4y² - 25 = (2y + 5)(2y - 5)

Example 3: Factorise 16x⁴ - 1 Step 1: 16x⁴ is a perfect square ((4x²)²) and 1 is a perfect square (1²).

Step 2: a = 4x², b = 1 Step 3: (a + b)(a - b)

Step 4: 16x⁴ - 1 = (4x² + 1)(4x² - 1). Notice that the second bracket is another DOTS and can be factorised further.

Step 5: 4x² - 1 = (2x + 1)(2x - 1)

Step 6: Final answer: 16x⁴ - 1 = (4x² + 1)(2x + 1)(2x - 1) 2.3 Factorising by Grouping in Pairs This technique is used when there are four or more terms and no common factor for all the terms.

Example 1: Factorise ax + ay + bx + by Step 1: Group the first two terms and the last two terms: (ax + ay) + (bx + by)

Step 2: Factor out the common factor from each group: a(x + y) + b(x + y)

Step 3: Notice that (x + y) is a common factor.

Factor it out: (x + y)(a + b) Therefore, ax + ay + bx + by = (x + y)(a + b)

Example 2: Factorise 2mp + 2mq - np - nq Step 1: (2mp + 2mq) + (-np - nq)

Step 2: 2m(p + q) - n(p + q)

Step 3: (p + q)(2m - n) Therefore, 2mp + 2mq - np - nq = (p + q)(2m - n)

Example 3: With rearranging: Factorise ab + 2b + a + 2 Step 1: (ab + a) + (2b + 2) (Direct grouping may not lead to a common bracket initially)

Step 2: a(b + 1) + 2(b + 1)

Step 3: (b + 1)(a + 2) Therefore, ab + 2b + a + 2 = (b+1)(a+2) Guided Practice (With Solutions)

Question 1: Expand and simplify: (3x - 2)(x + 5)

Solution: Step 1: 3x(x + 5) = 3x² + 15x Step 2: -2(x + 5) = -2x - 10 Step 3: (3x² + 15x) + (-2x - 10)

Step 4: 3x² + 15x - 2x - 10 = 3x² + 13x - 10 Comment: Remember to pay attention to the signs when multiplying. Combining like terms carefully is crucial for the final answer.

Question 2: Factorise completely: 9a² - 16b² Solution: Step 1: Recognise this as a difference of two squares.

Step 2: Identify a and b: a = 3a, b = 4b Step 3: Apply the formula: (a + b)(a - b)

Step 4: Substitute: (3a + 4b)(3a - 4b)

Comment: Identifying perfect squares is the key to spotting DOTS problems. Double-check your answer by expanding (3a + 4b)(3a - 4b) to see if you get back the original expression.

Question 3: Factorise completely: xy - 2x + 5y - 10 Solution: Step 1: Group in pairs: (xy - 2x) + (5y - 10)

Step 2: Factor out common factors: x(y - 2) + 5(y - 2)

Step 3: Factor out the common bracket: (y - 2)(x + 5)

Comment: Look for common factors within each group. The goal is to get the same expression inside the brackets so you can factor it out.

Question 4: Expand and simplify: (2p - 3)(p² + p - 1)

Solution: Step 1: 2p(p² + p - 1) = 2p³ + 2p² - 2p Step 2: -3(p² + p - 1) = -3p² - 3p + 3 Step 3: (2p³ + 2p² - 2p) + (-3p² - 3p + 3)

Step 4: 2p³ + 2p² - 3p² - 2p - 3p + 3 = 2p³ - p² - 5p + 3 Comment: Careful distribution and correct sign handling are essential. Take your time to combine like terms accurately.