Lesson Notes By Weeks and Term v5 - Grade 9
Algebraic expressions and factorisation (Grade 9) – Week 9 focus
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Algebraic expressions and factorisation (Grade 9) – Week 9 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 9
Term: 1st Term
Week: 9
Theme: General lesson support
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Algebraic expressions and factorisation are fundamental building blocks in mathematics. They provide a powerful language for representing and solving problems that arise in diverse real-life situations. This week, we will delve deeper into these concepts, focusing on advanced factorisation techniques and their application in simplifying expressions and solving equations. Understanding algebraic expressions and factorisation is not just about manipulating symbols; it's about developing critical thinking and problem-solving skills essential for success in future mathematics courses and various professions.
a.
Factorisation by Grouping: Sometimes, expressions don't immediately present a common factor for all terms. In such cases, we can group terms together to identify common factors within those groups, and then factorise further.
Why this works: By strategically grouping terms, we aim to create a common binomial factor that can then be factored out.
How to do it: Identify pairs of terms that share a common factor. Factor out the common factor from each pair. If the resulting expression has a common binomial factor, factor it out.
Example 1: Factorise `ax + ay + bx + by` Solution: Group the first two terms and the last two terms: `(ax + ay) + (bx + by)` Factor out 'a' from the first group and 'b' from the second group: `a(x + y) + b(x + y)` Now, notice that `(x + y)` is a common factor: `(x + y)(a + b)` Therefore, `ax + ay + bx + by = (x + y)(a + b)` Example 2: Factorise `2mn - 6m + n - 3` Solution: Group the first two terms and the last two terms: `(2mn - 6m) + (n - 3)` Factor out `2m` from the first group and `1` from the second group: `2m(n - 3) + 1(n - 3)` Now, notice that `(n - 3)` is a common factor: `(n - 3)(2m + 1)` Therefore, `2mn - 6m + n - 3 = (n - 3)(2m + 1)` b.
Difference of Two Squares (DOTS): This is a special case where we have an expression in the form a² - b². Its factorisation follows a specific pattern.
Why this works: The difference of squares is based on the expansion of (a + b)(a - b), which equals a² - b².
How to do it: Identify if the expression is in the form a² - b². This means both terms must be perfect squares, and they must be separated by a subtraction sign.
Apply the formula: `a² - b² = (a + b)(a - b)` Example 1: Factorise `x² - 9` Solution: Recognise that `x²` is a square of `x` and `9` is a square of `3`.
Apply the formula: `x² - 9 = (x + 3)(x - 3)` Example 2: Factorise `16p² - 25q²` Solution: Recognise that `16p²` is a square of `4p` and `25q²` is a square of `5q`.
Apply the formula: `16p² - 25q² = (4p + 5q)(4p - 5q)` c.
Factorising Quadratic Trinomials: Quadratic trinomials are expressions of the form `ax² + bx + c`, where a, b, and c are constants. We'll cover cases where a = 1 and a >
1. Case 1: a = 1 (Simple Trinomials)
Why this works: We're essentially reversing the process of expanding two binomials like (x + p)(x + q).
How to do it: Find two numbers that multiply to `c` and add up to `b`. Write the factors as `(x + p)(x + q)`, where `p` and `q` are the numbers you found in step
1. 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. Therefore, `x² + 5x + 6 = (x + 2)(x + 3)` Case 2: a > 1 (Harder Trinomials)
Why this works: Similar to the case where a = 1, but requires more careful manipulation to reverse the expansion of binomials.
How to do it (AC Method): Multiply `a` and `c` (AC). Find two numbers that multiply to AC and add up to `b`. Rewrite the middle term (bx) using the two numbers found in step
2. Factorise by grouping (as explained earlier).
Example 1: Factorise `2x² + 7x + 3` Solution: AC = 2 * 3 =
6. We need two numbers that multiply to 6 and add to
7. These numbers are 6 and
1. Rewrite the middle term: `2x² + 6x + x + 3` Factorise by grouping: `2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)` Therefore, `2x² + 7x + 3 = (x + 3)(2x + 1)` d.
Simplifying Algebraic Fractions: Simplifying algebraic fractions often involves factorising the numerator and/or denominator to identify common factors that can be cancelled.
Why this works: Just like with numerical fractions, simplifying by cancelling common factors results in an equivalent but simpler expression.
How to do it: Factorise the numerator and denominator as much as possible. Identify any common factors in the numerator and denominator. Cancel out the common factors.
Example 1: Simplify `(x² - 4) / (x + 2)` Solution: Factorise the numerator: `x² - 4 = (x + 2)(x - 2)` Rewrite the fraction: `((x + 2)(x - 2)) / (x + 2)` Cancel out the common factor `(x + 2)`: `(x - 2)` Therefore, `(x² - 4) / (x + 2) = x - 2` e. Solving Algebraic Equations by Factorisation: We can use factorisation to solve algebraic equations, especially quadratic equations.
Why this works: The zero-product property states that if the product of two or more factors is zero, then at least one of the factors must be zero.
How to do it: Rearrange the equation so that one side is equal to zero. Factorise the non-zero side of the equation. Set each factor equal to zero and solve for the variable.
Example 1: Solve `x² - 3x + 2 = 0` Solution: The equation is already in the form `ax² + bx + c = 0`.
Factorise the quadratic expression: `x² - 3x + 2 = (x - 1)(x - 2)` Set each factor equal to zero: `x - 1 = 0` or `x - 2 = 0` Solve for x: `x = 1` or `x = 2` Therefore, the solutions are `x = 1` and `x = 2`.