Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Factorise using the Highest Common Factor (HCF).
• Factorise a Difference of Two Squares (DOTS).
• Factorise trinomials in the form x² + bx + c.
• Simplify algebraic fractions by factorising.
• Solve simple equations by factorising.

Lesson summary

Algebraic expressions and factorisation are fundamental building blocks for further studies in mathematics. This week, we delve deeper into the techniques of manipulating and simplifying algebraic expressions through factorisation. Understanding factorisation allows us to solve equations, simplify complex problems, and model real-world situations effectively. In South Africa, understanding these concepts can help with budgeting, calculating profits in small businesses, and even optimising resource allocation in community projects. Without the ability to manipulate and simplify expressions, many everyday calculations would become excessively cumbersome.

Lesson notes

2.1 Highest Common Factor (HCF)

Factorisation: The HCF method involves identifying the largest factor common to all terms in an expression and then factoring it out. This is essentially the reverse of the distributive property. Why is this important? This simplifies expressions and makes them easier to work with. How does it work? Identify the HCF of the coefficients (the numbers in front of the variables). Identify the HCF of the variables (the highest power of each variable common to all terms). Write the HCF outside a set of brackets. Divide each term in the original expression by the HCF and write the result inside the brackets.

Example 1: Factorise `6x + 9`. HCF of 6 and 9 is

3. There is no variable common to both terms. `3( )` `6x / 3 = 2x` and `9 / 3 = 3`.

Therefore, `6x + 9 = 3(2x + 3)`.

Example 2: Factorise `12a²b - 18ab³`. HCF of 12 and 18 is

6. HCF of `a²b` and `ab³` is `ab`. `6ab( )` `12a²b / 6ab = 2a` and `-18ab³ / 6ab = -3b²`.

Therefore, `12a²b - 18ab³ = 6ab(2a - 3b²)`. 2.2 Difference of Two Squares (DOTS)

Factorisation: This method applies to expressions of the form `a² - b²`. It can be factorised as `(a + b)(a - b)`. Why is this important? It's a quick and efficient way to factorise certain quadratic expressions, especially useful in solving equations. How does it work? Ensure the expression is in the form of one square minus another square. Take the square root of each term.

Write two sets of brackets: one with a plus sign between the square roots, and one with a minus sign.

Example 3: Factorise `x² - 25`. `x²` is a square, and 25 is a square (5²). Square root of `x²` is x, and square root of 25 is

5. Therefore, `x² - 25 = (x + 5)(x - 5)`.

Example 4: Factorise `49m² - 64n²`. `49m²` is a square (7m)², and 64n² is a square (8n)². Square root of `49m²` is 7m, and square root of `64n²` is 8n.

Therefore, `49m² - 64n² = (7m + 8n)(7m - 8n)`. 2.3 Trinomial Factorisation (ax² + bx + c, where a = 1): This method is used to factorise expressions of the form `x² + bx + c`. Why is this important? Many quadratic equations can be expressed in this form, and factorising them allows us to find their solutions. How does it work? Find two numbers that multiply to give c (the constant term) and add up to give b (the coefficient of the x term). Write the factors as `(x + number 1)(x + number 2)`.

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

5. Those numbers are 2 and

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

Example 6: Factorise `x² - 7x + 12`. We need two numbers that multiply to 12 and add to -

7. Those numbers are -3 and -

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

Example 7: Factorise `x² + 2x - 15`. We need two numbers that multiply to -15 and add to

2. Those numbers are 5 and -

3. Therefore, `x² + 2x - 15 = (x + 5)(x - 3)`. 2.4 Simplifying Algebraic Fractions: This involves factorising the numerator and/or denominator and then cancelling any common factors. Why is this important? Simplifies complex fractions, making them easier to manipulate and solve. How does it work? Factorise the numerator and the denominator (if possible). Identify any common factors in the numerator and the denominator. Cancel the common factors.

Example 8: Simplify `(x² - 4) / (x + 2)`.

Factorise the numerator: `x² - 4 = (x + 2)(x - 2)`. The expression becomes `((x + 2)(x - 2)) / (x + 2)`. Cancel the common factor `(x + 2)`.

Therefore, `(x² - 4) / (x + 2) = x - 2`.

Example 9: Simplify `(x² + 3x + 2) / (x + 1)`.

Factorise the numerator: `x² + 3x + 2 = (x + 1)(x + 2)` The expression becomes `((x + 1)(x + 2))/(x + 1)` Cancel the common factor `(x + 1)` Therefore, `(x² + 3x + 2) / (x + 1) = x + 2` 2.5 Solving Equations by Factorisation: When an equation is set to equal zero, and one side can be factorised, we can use this to find the solutions. Why is this important? Efficiently finds the roots or solutions of many equations. How does it work? Rearrange the equation so that one side is equal to zero. Factorise the non-zero side. Set each factor equal to zero and solve for the variable.

Example 10: Solve for `x`: `x² + 4x + 3 = 0` Equation is already set to zero.

Factorise: `x² + 4x + 3 = (x + 1)(x + 3)` So, `(x + 1)(x + 3) = 0` Therefore, `x + 1 = 0` or `x + 3 = 0` Hence, `x = -1` or `x = -3` Guided Practice (With Solutions)

Question 1: Factorise `15p³q² + 25p²q`.

Solution: HCF of 15 and 25 is

5. HCF of `p³q²` and `p²q` is `p²q`. `5p²q( )` `15p³q² / 5p²q = 3pq` and `25p²q / 5p²q = 5`.

Therefore, `15p³q² + 25p²q = 5p²q(3pq + 5)`.

Commentary: This question reinforces HCF factorisation. Pay attention to the powers of the variables when finding the HC

F. Question 2: Factorise `16m² - 81`.

Solution: `16m²` is a square (4m)² and 81 is a square (9)². Square root of `16m²` is 4m and square root of 81 is

9. Therefore, `16m² - 81 = (4m + 9)(4m - 9)`.

Commentary: This question tests the understanding of DOTS. Remember to check if both terms are perfect squares and are separated by a subtraction sign.