Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Factorise using difference of two squares.
• Factorise trinomials.
• Simplify algebraic fractions.
• Apply factorisation to solve simple equations.
• Identify common errors in factorisation.

Lesson summary

This week, we delve deeper into the world of algebraic expressions and factorisation. Algebraic expressions are the building blocks of algebra, and factorisation is like reverse engineering – breaking down complex expressions into simpler ones. Understanding these concepts is crucial not just for your mathematics marks, but also for problem-solving in various real-life scenarios, from budgeting to understanding scientific formulas. In South Africa, where resource management is key, these skills are highly valuable.

Lesson notes

2. 1. Difference of Two Squares The difference of two squares is a special algebraic expression in the form: a² - b².

It can be factorised as follows: a² - b² = (a + b)(a - b) Why? This pattern arises from expanding (a + b)(a - b): (a + b)(a - b) = a² - ab + ab - b² = a² - b² How? Identify two terms that are perfect squares separated by a minus sign. Then, take the square root of each term and apply the formula.

Example 1: Factorise x² - 9 Solution: x² is a perfect square (√x² = x) and 9 is a perfect square (√9 = 3).

Therefore, x² - 9 = (x + 3)(x - 3)

Example 2: Factorise 4y² - 25 Solution: 4y² is a perfect square (√(4y²) = 2y) and 25 is a perfect square (√25 = 5).

Therefore, 4y² - 25 = (2y + 5)(2y - 5)

Example 3: Factorise 16a² - 81b² Solution: 16a² is a perfect square (√(16a²) = 4a) and 81b² is a perfect square (√(81b²) = 9b).

Therefore, 16a² - 81b² = (4a + 9b)(4a - 9b) 2.

2. Factorising Trinomials Trinomials are algebraic expressions with three terms. We will focus on factorising quadratic trinomials in the form: ax² + bx + c. In Grade 9, we often deal with cases where a = 1, i.e., x² + bx + c. Factorising x² + bx + c: How?

Find two numbers that: Multiply to give 'c' Add to give 'b' If these two numbers are p and q, then the trinomial factorises to: (x + p)(x + q) Why? Expanding (x + p)(x + q) gives: x² + px + qx + pq = x² + (p + q)x + pq. So, (p + q) must equal 'b' and pq must equal 'c'.

Example 1: Factorise x² + 5x + 6 Solution: We need two numbers that multiply to 6 and add to

5. These numbers are 2 and 3 (2 x 3 = 6 and 2 + 3 = 5).

Therefore, x² + 5x + 6 = (x + 2)(x + 3)

Example 2: Factorise x² - 7x + 12 Solution: We need two numbers that multiply to 12 and add to -

7. These numbers are -3 and -4 (-3 x -4 = 12 and -3 + -4 = -7).

Therefore, x² - 7x + 12 = (x - 3)(x - 4)

Example 3: Factorise x² + 2x - 15 Solution: We need two numbers that multiply to -15 and add to

2. These numbers are 5 and -3 (5 x -3 = -15 and 5 + -3 = 2).

Therefore, x² + 2x - 15 = (x + 5)(x - 3) 2.

3. Perfect Square Trinomials A perfect square trinomial is a trinomial that results from squaring a binomial.

They have the form: a² + 2ab + b² = (a + b)² a² - 2ab + b² = (a - b)² How to Identify? The first and last terms are perfect squares. The middle term is twice the product of the square roots of the first and last terms.

Example 1: Factorise x² + 6x + 9 Solution: x² is a perfect square (√x² = x), 9 is a perfect square (√9 = 3). 2 x x x 3 = 6x (the middle term).

Therefore, x² + 6x + 9 = (x + 3)² Example 2: Factorise 4y² - 20y + 25 Solution: 4y² is a perfect square (√(4y²) = 2y), 25 is a perfect square (√25 = 5). 2 x 2y x 5 = 20y (the middle term, noting the minus sign).

Therefore, 4y² - 20y + 25 = (2y - 5)² 2.

4. Simplifying Algebraic Fractions Simplifying algebraic fractions involves factorising the numerator and/or the denominator, then cancelling any common factors. How? Factorise the numerator and denominator completely. Identify any common factors between the numerator and the denominator. Cancel out the common factors.

Example 1: Simplify (x² - 4) / (x + 2)

Solution: Factorise the numerator: x² - 4 = (x + 2)(x - 2)

The fraction becomes: [(x + 2)(x - 2)] / (x + 2) Cancel the common factor (x + 2): (x - 2) Therefore, (x² - 4) / (x + 2) = x - 2 Example 2: Simplify (x² + 3x + 2) / (x² + 5x + 6)

Solution: Factorise the numerator: x² + 3x + 2 = (x + 1)(x + 2)

Factorise the denominator: x² + 5x + 6 = (x + 2)(x + 3)

The fraction becomes: [(x + 1)(x + 2)] / [(x + 2)(x + 3)] Cancel the common factor (x + 2): (x + 1) / (x + 3) Therefore, (x² + 3x + 2) / (x² + 5x + 6) = (x + 1) / (x + 3) Guided Practice (With Solutions)

Question 1: Factorise fully: 9a² - 16 Solution: This is a difference of two squares. 9a² is a perfect square (√(9a²) = 3a) and 16 is a perfect square (√16 = 4).

Therefore, 9a² - 16 = (3a + 4)(3a - 4).

Commentary: We directly applied the difference of two squares formula.

Question 2: Factorise fully: x² - 8x + 16 Solution: This is a trinomial. Let's check if it's a perfect square trinomial. x² is a perfect square (√x² = x), 16 is a perfect square (√16 = 4). 2 x 4 = 8x. Since the middle term is -8x, it's a perfect square trinomial.

Therefore, x² - 8x + 16 = (x - 4)².

Commentary: Recognizing the perfect square trinomial pattern saves time.

Question 3: Factorise fully: m² + 5m - 14 Solution: This is a trinomial. We need two numbers that multiply to -14 and add to

5. The numbers are 7 and -2 (7 * -2 = -14, 7 + -2 = 5).

Therefore, m² + 5m - 14 = (m + 7)(m - 2).

Commentary: Careful attention to the signs is crucial.

Question 4: Simplify: (y² - 9) / (y - 3)

Solution: First, factorise the numerator: y² - 9 = (y + 3)(y - 3)

The fraction becomes: [(y + 3)(y - 3)] / (y - 3)

Cancel the common factor (y - 3): (y + 3) Therefore, (y² - 9) / (y - 3) = y +

3. Commentary: Factorising is the key first step.