Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Factorise quadratic trinomials (ax² + bx + c, where a=1).
• Factorise the difference of two squares (DOTS).
• Factorise by taking out a common factor.
• Simplify algebraic fractions using factorisation.

Lesson summary

This week, we delve deeper into the fascinating world of algebraic expressions and factorisation. Building upon your Grade 8 foundation, we'll explore more advanced techniques to manipulate and simplify algebraic expressions. This skill is crucial, not just for future maths studies, but also for everyday problem-solving. Imagine budgeting your monthly allowance, calculating the cost of building a small structure, or even understanding interest rates on a loan – algebraic thinking is at the heart of all these situations.

Lesson notes

2.1 Factorisation: Unpacking Algebraic Expressions Factorisation is the reverse process of expansion (multiplying out brackets). It involves expressing an algebraic expression as a product of its factors. Think of it like finding the ingredients that were mixed together to create a specific dish. These "ingredients" are the factors. Understanding factorisation allows us to simplify expressions, solve equations, and gain deeper insights into the relationships between variables. 2.2 Common Factor Factorisation This is the most fundamental type of factorisation. We look for the greatest common factor (GCF) of all the terms in the expression and then factor it out.

Example 1: Factorise 3x + 6 The GCF of 3x and 6 is

3. Therefore, 3x + 6 = 3(x + 2) Why?* We divide each term by the GCF (3). 3x/3 = x and 6/3 =

2. The '3' is then placed outside the brackets. How?* Always check your answer by expanding: 3(x+2) = 3x +

6. This verifies your factorisation.

Example 2: Factorise 4a²b - 8ab² The GCF of 4a²b and 8ab² is 4ab.

Therefore, 4a²b - 8ab² = 4ab(a - 2b) Why? 4ab is the largest expression that divides evenly into both terms.

Think of breaking down each term: 4a²b = 4 a a b and 8ab² = 4 2 a b b. The common parts are 4 a b. How?* Check by expanding: 4ab(a - 2b) = 4a²b - 8ab². 2.3 Difference of Two Squares (DOTS) This pattern arises when we have the difference between two perfect squares. The general form is a² - b² = (a + b)(a - b).

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

Therefore, x² - 9 = (x + 3)(x - 3) Why?* This pattern comes directly from expanding (a+b)(a-b): (a+b)(a-b) = a² - ab + ab - b² = a² - b². How?* Check by expanding: (x + 3)(x - 3) = x² - 3x + 3x - 9 = x² -

9. Example 2: Factorise 16p² - 25q² 16p² is a perfect square (4p 4p) and 25q² is a perfect square (5q * 5q).

Therefore, 16p² - 25q² = (4p + 5q)(4p - 5q) Why?* Again, this is a direct application of the DOTS pattern. How?* Check by expanding: (4p + 5q)(4p - 5q) = 16p² - 20pq + 20pq - 25q² = 16p² - 25q². 2.4 Factorising Quadratic Trinomials (ax² + bx + c, where a = 1) This involves breaking down a quadratic expression into the product of two binomials.

Example 1: Factorise x² + 5x + 6 We need to find two numbers that add up to 5 (the coefficient of the x term) and multiply to give 6 (the constant term). The numbers are 2 and 3 (2 + 3 = 5 and 2 3 = 6).

Therefore, x² + 5x + 6 = (x + 2)(x + 3) Why? This works because when we expand (x+2)(x+3), we get x² + 3x + 2x + 6 = x² + 5x +

6. The 3x and 2x are where our 'adding to 5' comes from, and the 23 gives us the '6'. How?* Check by expanding: (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6 Example 2: Factorise x² - 2x - 8 We need two numbers that add up to -2 and multiply to give -

8. The numbers are -4 and 2 (-4 + 2 = -2 and -4 2 = -8).

Therefore, x² - 2x - 8 = (x - 4)(x + 2) Why?* The same logic applies as in the previous example. The signs are crucial here! How?* Check by expanding: (x - 4)(x + 2) = x² + 2x - 4x - 8 = x² - 2x - 8 2.5 Simplifying Algebraic Fractions using Factorisation We can often simplify algebraic fractions by factorising the numerator and/or denominator and then cancelling common factors.

Example: Simplify (x² + 4x + 3) / (x + 1) First, factorise the numerator: x² + 4x + 3 = (x + 1)(x + 3)

Now the fraction is: (x + 1)(x + 3) / (x + 1) We can cancel the common factor of (x + 1) Therefore, the simplified expression is (x + 3) Why?* Cancelling common factors is the same as dividing both numerator and denominator by the same thing, which doesn't change the value of the fraction (as long as the factor is not zero). How?* Substitute a value for x into both the original and simplified expressions. If they give the same answer (avoiding x = -1!), you've simplified correctly. Guided Practice (With Solutions)

Question 1: Factorise 5x² + 10x.

Solution: The GCF of 5x² and 10x is 5x. 5x² + 10x = 5x(x + 2)

Commentary:* This is a straightforward application of common factor factorisation. Identifying the GCF is key.

Question 2: Factorise a² -

4

9. Solution: This is a difference of two squares. a² is a perfect square, and 49 is a perfect square (7 7). a² - 49 = (a + 7)(a - 7)

Commentary:* Recognising the DOTS pattern is crucial.

Remember the formula: a² - b² = (a + b)(a - b).

Question 3: Factorise x² + 8x + 15 Solution: We need two numbers that add up to 8 and multiply to give

1

5. The numbers are 3 and 5 (3 + 5 = 8 and 3 5 = 15). x² + 8x + 15 = (x + 3)(x + 5)

Commentary:* Practice finding the correct number pairs. Start by listing factor pairs of the constant term (15).

Question 4: Simplify (x² - 16) / (x + 4)

Solution: Factorise the numerator (x² - 16): This is DOTS, so x² - 16 = (x + 4)(x - 4)

Now the fraction is: (x + 4)(x - 4) / (x + 4) Cancel the common factor of (x + 4) The simplified expression is (x - 4)

Commentary:* This combines DOTS factorisation with simplifying algebraic fractions.