Comprehensive Study Notes: Expanding and Factorising Linear Expressions

Unit Learning Goal and Success Criteria

Learning Goal 9 focuses on the application of knowledge regarding algebraic properties to expand and factorise linear expressions. By the completion of this unit, the following Success Criteria should be met:

• 9.1: Define the distributive law and understand how it can be used to expand brackets.
• 9.2: Expand an algebraic expression using the distributive law.
• 9.3: Expand and simplify an algebraic expression by using the distributive law and by adding and subtracting like terms.
• 9.4: Define factorise.
• 9.5: Recall the Highest Common Factor (HCF).
• 9.6: Identify the Highest Common Factor (HCF) within two or more terms.
• 9.7: Factorise an expression by using the Highest Common Factor (HCF).
• 9.8: Apply algebraic properties to write and evaluate algebraic expressions to model different situations.

Understanding Expansion and the Distributive Law

Expansion of brackets is a process based on the principle that two expressions can be equal in value while looking different in form. For example, $2(x+2) = 2x+4$. Expanding brackets helps rewrite expressions to combine like terms and find unknown values more efficiently. This is achieved using the distributive law.

The basic terminology of algebra relevant to this unit includes:

• Pronumerals (Variables): Symbols (usually letters) used to represent unknown values.
• Expression: A collection of terms and operators without an equals sign.
• Term: A single part of an expression, which can be a number, a variable, or a product of both.
• Coefficient: The numerical factor of a term containing a variable.
• Constant Term: A term that consists of only a number.

Procedures for Expanding Expressions

To expand an expression such as $a(b+c)$, the distributive law is applied by multiplying the term outside the brackets by each term inside the brackets: $a \times b + a \times c$.

Example 1: Expanding basic terms

• $3(2x+5) = 3(2x) + 3(5) = 6x+15$
• $4x(2-y) = 4x(2) - 4x(y) = 8x-4xy$
• $-8(7+2y) = -8(7) + (-8)(2y) = -56 - 16y$

Alternative Layout for Expansion: An alternative visual layout uses a grid-like structure to multiply terms: For $3(2x+5)$, one can place $2x$ and $+5$ at the top of a table and $3$ on the side:

• $3 \times 2x = 6x$
• $3 \times 5 = 15$
• Result: $6x+15$

Example 2: We Do Practice

• $5(3x+4) = 5(3x) + 5(4) = 15x+20$
• $7a(4-b) = 7a(4) - 7a(b) = 28a - 7ab$
• $-2(8+5b) = -2(8) + (-2)(5b) = -16 - 10b$

Expanding and Collecting Like Terms

When an expression involves an expansion followed by additional terms, the distributive law is used first, and then the expression is simplified by adding or subtracting like terms.

Example Calculations:

• $5(3x+2)+2 = 15x+10+2 = 15x+12$ (Note: The transcript contains various versions of this calculation; for instance: $15+10+2 = 17+10$ which appears as a mid-step arithmetic error or specific numerical example in the source slides).
• $5+3(10-x) = 5+3(10)-3(x) = 5+30-3x = 35-3x$

Define and Perform Factorisation

Factorising is defined as the opposite procedure to expanding. It allows for the simplification of expressions and the resolution of more complex mathematical problems. For instance, $2x+4$ is factorised to $2(x+2)$.

The Highest Common Factor (HCF): The HCF of a set of terms is the largest factor that divides into each term. Determining the HCF is the first step in factorising an expression.

• HCF of $15x$ and $21y$ is $3$.
• HCF of $10a$ and $20c$ is $10$.
• HCF of $12x$ and $18xy$ is $6x$.

Steps to Factorise:

1. Identify the HCF of the terms in the expression.
2. Place the HCF outside the brackets.
3. Divide each term by the HCF and record the result inside the brackets.
4. Check the answer by expanding the factorised form to see if it returns the original expression.

Examples of Factorisation:

• $6x+15$: HCF is $3$. Dividing each term: $6x \div 3 = 2x$ and $15 \div 3 = 5$. Result: $3(2x+5)$.
• $12a+18ab$: HCF is $6a$. Dividing each term: $12a \div 6a = 2$ and $18ab \div 6a = 3b$. Result: $6a(2+3b)$.
• $21x-14y$: HCF is $7$. Dividing each term: $21x \div 7 = 3x$ and $14y \div 7 = 2y$. Result: $7(3x-2y)$.

Exercise 5G: Expanding Brackets and Problem Solving

Fluency Exercises

1. Expand using the distributive law:    * a. $5(3x+2) = 15x+10$    * b. $7(2x+1) = 14x+7$    * c. $6(3x+5) = 18x+30$    * d. $10(4x+3) = 40x+30$
2. Further expansion practice:    * a. $3(2a+5) = 6a+15$    * b. $5(3t+4) = 15t+20$    * c. $8(2m+4) = 16m+32$    * d. $3(v+6) = 3v+18$    * e. $4(3-2j) = 12-8j$    * f. $6(2k-5) = 12k-30$    * g. $4(3m-1) = 12m-4$    * h. $2(8-c) = 16-2c$
3. Expansion with negative integers and variables:    * a. $-5(9+g) = -45-5g$    * b. $-7(5b+4) = -35b-28$    * c. $-9(u-9) = -9u+81$    * d. $-8(5-h) = -40+8h$    * e. $8z(k-h) = 8zk-8zh$    * f. $-6j(k+a) = -6jk-6ja$    * g. $4u(2r-q) = 8ur-4uq$    * h. $4m(5w-3a) = 20mw-12ma$

Problem-Solving

7. Write and expand expressions for:    * a. A number $t$ has $4$ added to it and the result is multiplied by $3$. Expression: $3(t+4) = 3t+12$.    * b. A number $u$ has $3$ subtracted from it and the result is doubled. Expression: $2(u-3) = 2u-6$.    * c. A number $v$ is doubled, then $5$ is added. The result is tripled. Expression: $3(2v+5) = 6v+15$.    * d. A number $w$ is tripled, then $2$ is subtracted. The result is doubled. Expression: $2(3w-2) = 6w-4$.
8. Operations matching:    * a. $x$ is doubled and $6$ is added ($2x+6$) matches D: $x$ is increased by $3$ and the result is doubled ($2(x+3) = 2x+6$).    * b. $x$ is reduced by $5$ and result doubled ($2(x-5)$) matches A: $x$ is doubled and reduced by $10$ ($2x-10$).    * c. $x$ is added to double the value of $x$ ($2x+x$) matches B: The number $x$ is tripled ($3x$).    * d. $x$ is halved, then $3$ is added and the result doubled ($2(\frac{x}{2} + 3)$) matches E: $x$ is increased by $6$ ($x+6$).    * e. $2$ is subtracted from one-third of $x$ and result tripled ($3(\frac{x}{3} - 2)$) matches C: $x$ is decreased by $6$ ($x-6$).
9. Classroom Context:    * a. Total pencils for $s$ students and $t$ teachers: $5s+3t$.    * b. Cost at $\$2$ each: $2(5s+3t) = 10s+6t$.    * c. Pencils and cases (cases cost $\$4$ each): Total cases cost $4(s+t) = 4s+4t$. Total cost = $(10s+6t) + (4s+4t) = 14s+10t$.    * d. For $s=20, t=4$: $14(20) + 10(4) = 280+40 = \$320$.

Reasoning

11. Prove $4a(2+b)+2ab$ is equivalent to $a(6b+4)+4a$:    * Left side: $8a+4ab+2ab = 8a+6ab$.    * Right side: $6ab+4a+4a = 8a+6ab$.    * Both are equivalent.
12. Area Model for $(2a+3b)(7+c)$. The grid shows:    * Row 1: $7 \times 2a = 14a$; $7 \times 3b = 21b$.    * Row 2: $c \times 2a = 2ac$; $c \times 3b = 3bc$.    * Sum: $14a+21b+2ac+3bc$.
13. Mental Math with Distributive Law:    * a. $9 \times 204 = 9(200+4) = 1800+36 = 1836$.    * b. $204 \times 9 = 204(10-1) = 2040-204 = 1836$.    * c. $\times 11$ rule ($10a+a$):      * i. $14 \times 11 = 140+14 = 154$      * ii. $32 \times 11 = 320+32 = 352$      * iii. $57 \times 11 = 570+57 = 627$      * iv. $79 \times 11 = 790+79 = 869$

Real-World Applications of Algebraic Modeling

Algebraic properties are applicable in engineering, sciences, and economics to model specific situations.

Case Study: Hourly Rates and Service Costs

• If an hourly rate is $\$12$, the cost ($C$) for $h$ hours is $C = 12h$.
• For a service that has a call-out fee of $\$35$ and an hourly rate of $\$80$, the expression for total cost is $80h+35$.
• Perimeter of a rectangle where height $H = h$ and width $W = 3+h$: $P = 2(W+H) = 2((3+h)+h) = 2(2h+3) = 4h+6$.

Pricing and Deals:

• Deal 1: Cost = $10+4n$.
• Deal 2: Cost = $20+n$.
• Deal 3: Evaluating for specific quantities (e.g., $n=3, 4.10, 10$ students/hours) determines which deal is most cost-effective.

Geometric Reasoning Case:

• Expression for the perimeter of a shape with dimensions $x+2$ and height/width logic: $2(x+2) = 2x+4$.
• If dimensions resulted in negative values (e.g., side lengths like $x=-6$ in an expression $x+5$), the shape cannot exist physically.

Summary of Factorisation Exercises and Solutions (9.6 & 9.7)

• Page 31 Solutions:     * a. $6x+15 = 3(2x+5)$     * b. $12a+18ab = 6a(2+3b)$     * c. $21x-14y = 7(3x-2y)$
• Page 33 Solutions:     * a. $12x+30 = 6(2x+5)$. HCF of $12$ and $30$ is $6$.     * b. $15a+25ab = 5a(3+5b)$. HCF of $15$ and $25$ is $5$, plus variable $a$.     * c. $18x-15y = 3(6x-5y)$. HCF of $18$ and $15$ is $3$.
• Page 54 Solutions (Modeling):     * a. i. $2(x+2)$, ii. $2x+4$, iii. $4(x+1)$. All expand to $2x+4$.     * Factorisation can be used to compare different pricing models (Deals A through F). For instance, Deal A was the choice for $86$ buyers, whereas Deal F was selected by only $3$ people.