Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Factorise using a common factor (HCF).
• Factorise a difference of two squares.
• Factorise by grouping terms.
• Factorise trinomials (where a=1).
• Simplify algebraic fractions using factorisation.

Lesson summary

Algebraic expressions and factorization are fundamental concepts in mathematics, forming the bedrock for more advanced topics like solving equations, graphing functions, and calculus. Mastering these skills is crucial for success in subsequent mathematics courses and has practical applications in various fields. In a South African context, understanding algebraic expressions can help with budgeting, calculating costs for small businesses, understanding financial investments, and even interpreting statistical data related to social and economic trends. This week, we focus on consolidating skills in factorizing various types of expressions.

Lesson notes

2.1 Factorising by Common Factor (HCF) Factorising is the reverse process of expanding brackets. When we factorise, we look for the highest common factor (HCF) of all the terms in the expression and write the expression as a product of the HCF and the remaining factor.

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

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

Explanation: We divide each term by the HCF (3) and write the result inside the bracket.

Example 2: Factorise 12a²b - 18ab². The HCF of 12a²b and 18ab² is 6ab.

Therefore, 12a²b - 18ab² = 6ab(2a - 3b).

Explanation: Again, we identified the HCF of both the coefficients and the variable parts and divide each term to find the expression to place inside the bracket. 2.2 Factorising the Difference of Two Squares The difference of two squares is an expression of the form a² - b².

It can be factorised as follows: a² - b² = (a + b)(a - b)

Example 1: Factorise x² -

1

6. Here, x² is a square and 16 = 4² is a square.

Therefore, x² - 16 = (x + 4)(x - 4).

Explanation: We recognized the expression as a difference of two squares and applied the standard formula.

Example 2: Factorise 49y² - 25z². Here, 49y² = (7y)² and 25z² = (5z)².

Therefore, 49y² - 25z² = (7y + 5z)(7y - 5z).

Explanation: Similar to example 1, we first express each term as a square and then apply the difference of two squares formula. 2.3 Factorising by Grouping Factorising by grouping is used when there is no single common factor for all terms in the expression. In this method, we group terms together and find the common factor for each group.

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

Factorise each group: a(x + y) + b(x + y). Now, (x + y) is a common factor: (x + y)(a + b).

Therefore, ax + ay + bx + by = (x + y)(a + b).

Explanation: We arranged the terms and then successfully extracted common factors until we could rewrite the entire expression.

Example 2: Factorise 2mp - 2nq + 3mq - 3np.

Rearrange the terms: 2mp + 3mq - 2nq - 3np.

Group the terms: (2mp + 3mq) + (-2nq - 3np).

Factorise each group: m(2p + 3q) - n(2q + 3p). Notice that (2p + 3q) = (3p + 2q) is not the same.

Rearrange again: 2mp - 3np - 2nq + 3mq Group the terms: (2mp - 3np) + (-2nq + 3mq).

Factorise each group: p(2m - 3n) + q(-2n + 3m).

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

Explanation: In this case, careful re-arrangement of terms was key to being able to then factorise using the common factor method. 2.4 Factorising Trinomials (ax² + bx + c, where a = 1) A trinomial is an expression with three terms. A trinomial of the form x² + bx + c can be factorised by finding two numbers that add up to 'b' and multiply to 'c'.

Example 1: Factorise x² + 5x +

6. We need two numbers that add up to 5 and multiply to

6. These numbers are 2 and

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

Explanation: We identified the two numbers which satisfied the conditions and wrote them in the required form.

Example 2: Factorise x² - 2x -

8. We need two numbers that add up to -2 and multiply to -

8. These numbers are -4 and

2. Therefore, x² - 2x - 8 = (x - 4)(x + 2).

Explanation: Here, one of the numbers had to be negative for the multiplication to result in a negative value, which affected the addition condition too. Guided Practice (With Solutions)

Question 1: Factorise 15x³y² + 25x²y³.

Solution: The HCF of 15x³y² and 25x²y³ is 5x²y². 15x³y² + 25x²y³ = 5x²y²(3x + 5y).

Commentary: First, identify the HCF (numerical and variable parts). Then, divide each term by the HCF and write the result within the bracket.

Question 2: Factorise 81 - y².

Solution: 81 - y² = 9² - y² = (9 + y)(9 - y).

Commentary: Recognize the expression as a difference of two squares. Apply the formula a² - b² = (a + b)(a - b).

Question 3: Factorise 3ax + 3ay - bx - by.

Solution: Group the terms: (3ax + 3ay) + (-bx - by).

Factorise each group: 3a(x + y) - b(x + y). (x + y)(3a - b).

Therefore, 3ax + 3ay - bx - by = (x + y)(3a - b).

Commentary: This demonstrates factorisation by grouping. Pay attention to the signs when factoring out the second group.

Question 4: Factorise x² + 8x +

1

5. Solution: We need two numbers that add up to 8 and multiply to

1

5. These numbers are 3 and

5. Therefore, x² + 8x + 15 = (x + 3)(x + 5).

Commentary: Finding the correct factors is crucial. In this case, both numbers had to be positive.

Question 5: Simplify the fraction: (x² - 4) / (x + 2)

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

The expression now becomes: ( (x + 2)(x - 2) ) / (x + 2) Cancel out the common factor (x + 2)

The simplified fraction is: x - 2

Commentary: A key skill is to use factorisation to help simplify fractions, which has many uses.