MYP9 Unit 3: Algebraic Expressions and Formulae

1. Expansion Rules

The Distributive Law

The distributive law is a fundamental principle in algebra that allows us to multiply a single term by each term inside a set of parentheses.

Definition:
Consider the expression a(b + c). According to the distributive law, this can be expanded to:

a(b + c) = ab + ac

Example 1:

Expand the following expressions:

1. 3(4x + 1)
2. 2x(5 - 2x)
3. -2x(x - 3)

Solutions:

1. 3(4x + 1) = 3 × 4x + 3 × 1 = 12x + 3
2. 2x(5 - 2x) = 2x × 5 - 2x × 2x = 10x - 4x²
3. -2x(x - 3) = -2x × x + (-2x) × (-3) = -2x² + 6x

The Product of binomials

When expanding products of binomials, we use the distributive law multiple times. One common method for expanding binomials is the FOIL method, which stands for First, Outer, Inner, Last.

FOIL Method:

Consider the product (a + b)(c + d). The FOIL method involves multiplying:

• First terms: a × c
• Outer terms: a × d
• Inner terms: b × c
• Last terms: b × d

Then, sum these products to get the expanded form.

(a + b)(c + d) = ac + ad + bc + bd

Example 2:

Expand and simplify:

1. (x + 3)(x + 2)
2. (2x + 1)(3x - 2)

Solutions:

1. Using the FOIL method:
   • First: x × x = x²
   • Outer: x × 2 = 2x
   • Inner: 3 × x = 3x
   • Last: 3 × 2 = 6

   Combine: x² + 2x + 3x + 6 = x² + 5x + 6

2. Using the FOIL method:
   • First: 2x × 3x = 6x²
   • Outer: 2x × -2 = -4x
   • Inner: 1 × 3x = 3x
   • Last: 1 × -2 = -2

   Combine: 6x² - 4x + 3x - 2 = 6x² - x - 2

Difference of Two Squares

The difference of two squares is a specific form that can be factored into the product of two binomials.

Definition:
For any two terms a and b:

a² - b² = (a + b)(a - b)

Example 3:

Expand and simplify:

1. (x + 5)(x - 5)
2. (3x + 4y)(3x - 4y)

Solutions:

1. (x + 5)(x - 5) = x² - 25
2. (3x + 4y)(3x - 4y) = (3x)² - (4y)² = 9x² - 16y²

Perfect Squares Expansion

The perfect square expansion formula is used to expand expressions where the binomial is squared.

Definition:
For any binomial a + b:

(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²

Example 4:

Expand and simplify:

1. (5x + 1)²
2. (4 - 3x)²

Solutions:

1. (5x + 1)² = (5x)² + 2 × 5x × 1 + 1² = 25x² + 10x + 1
2. (4 - 3x)² = (4)² - 2 × 4 × 3x + (3x)² = 16 - 24x + 9x²

Further Expansion

For expressions involving more complex terms, the distributive law is applied iteratively.

Example 5:

Expand and simplify:

1. (2x + 3)(x² + 4x + 5)

Solution:

1. (2x + 3)(x² + 4x + 5) = 2x(x² + 4x + 5) + 3(x² + 4x + 5) = 2x³ + 8x² + 10x + 3x² + 12x + 15 = 2x³ + 11x² + 22x + 15

Summary of Expansion Formulas

1. Distributive Law:
   a(b + c) = ab + ac
2. FOIL Method for Binomials:
   (a + b)(c + d) = ac + ad + bc + bd
3. Difference of Two Squares:
   a² - b² = (a + b)(a - b)
4. Perfect Squares:
   (a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²

2. Factorisation by Taking Out Common Factor

Explanation

Factorisation is the process of breaking down an expression into a product of simpler factors. One common method of factorisation is by taking out the common factor from the terms of the expression. This method involves identifying the greatest common factor (GCF) of all the terms in the expression and factoring it out.

Definition:
If every term in an expression has a common factor, this factor can be placed in front of a set of parentheses, with the remaining terms inside the parentheses. This is essentially the reverse of the distributive law.

Example 1:

Factorise the following expressions:

1. 6x² + 4x
2. 15y³ - 9y²
3. -4(a + 1) + (a + 2)(a + 1)

Solutions:

1. 6x² + 4x
   • Identify the GCF: 2x
   • Factor out the GCF:
     6x² + 4x = 2x(3x + 2)
2. 15y³ - 9y²
   • Identify the GCF: 3y²
   • Factor out the GCF:
     15y³ - 9y² = 3y²(5y - 3)
3. -4(a + 1) + (a + 2)(a + 1)
   • Identify the common factor: (a + 1)
   • Factor out the common factor:
     -4(a + 1) + (a + 2)(a + 1) = (a + 1)(-4 + (a + 2)) = (a + 1)(a - 2)

Example 2:

Factorise the following expression:

8x³ + 4x² - 2x

Solution:

• Identify the GCF: 2x
• Factor out the GCF:
  8x³ + 4x² - 2x = 2x(4x² + 2x - 1)

3. Factorisation by Grouping

Explanation

Factorisation by grouping is a method used to factor polynomials that do not have a single common factor for all terms. This method involves rearranging and grouping terms in such a way that each group has a common factor. By factoring out the common factor from each group, the expression can be simplified further.

Factorisation by grouping involves the following steps:

1. Arrange the terms into groups.
2. Factor out the common factor from each group.
3. If the resulting binomials have a common factor, factor out the common binomial factor.

Example:

Factorise the following expression:

x³ + 3x² + 2x + 6

Solution:

• Group the terms:
  (x³ + 3x²) + (2x + 6)
• Factor out the common factor from each group:
  x²(x + 3) + 2(x + 3)
• Factor out the common binomial factor (x + 3):
  (x + 3)(x² + 2)

4. Factorisation of Identities

Explanation

Algebraic identities are equations that hold true for all values of the variables involved. These identities can be used to simplify expressions through factorisation. Two of the most common algebraic identities used in factorisation are the square of a binomial and the difference of squares.

Common Algebraic Identities:

• Square of a Binomial:
  • \((a + b)^2 = a^2 + 2ab + b^2\)
  • \((a – b)^2 = a^2 – 2ab + b^2\)
• Difference of Squares:
  • \(a^2 – b^2 = (a + b)(a – b)\)

Factorisation Using Identities

Example 1:

Factorise the following expression using the square of a binomial identity:

4x² + 12x + 9

Solution:

Recognize the form \(a² + 2ab + b² = (a + b)²\):

\(4x² + 12x + 9 = (2x)² + 2(2x)(3) + 3²
= (2x + 3)²\)

Example 2:

Factorise the following expression using the difference of squares identity:

16y² - 25

Solution:

Recognize the form \(a² – b² = (a + b)(a – b)\):

\(16y² – 25 = (4y)² – 5²
= (4y + 5)(4y – 5)\)

5. Factorisation of Trinomials

Explanation

Factorising trinomials is a method used to express a trinomial (a polynomial with three terms) as a product of two binomials. This method is particularly useful for solving quadratic equations. There are different approaches to factorising trinomials, but one common method involves splitting the middle term.

Definition:
A trinomial is an algebraic expression of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. The goal is to express this trinomial as the product of two binomials: \((dx + e)(fx + g)\).

Factorising Trinomials When \(a = 1\)

When the leading coefficient \(a\) is 1, the trinomial takes the form \(x^2 + bx + c\). The steps for factorisation are simpler in this case.

Steps:

1. Find two numbers that multiply to \(c\) and add up to \(b\).
2. Use these two numbers to split the middle term \(bx\).
3. Factor by grouping.

Example 1:

Factorise the trinomial:

x² - 7x + 12

Solution:

1. Find two numbers that multiply to 12 and add up to -7: -3 and -4.
2. Split the middle term using these numbers:
   x² - 3x - 4x + 12
3. Group the terms into pairs:
   (x² - 3x) - (4x - 12)
4. Factor out the common factor from each pair:
   x(x - 3) - 4(x - 3)
5. Factor out the common binomial factor:
   (x - 4)(x - 3)

Factorising Trinomials When \(a \neq 1\)

When the leading coefficient \(a\) is not 1, the trinomial takes the form \(ax^2 + bx + c\). The factorisation process involves a few additional steps.

Steps:

1. Multiply the leading coefficient \(a\) by the constant term \(c\).
2. Find two numbers that multiply to \(ac\) and add up to \(b\).
3. Split the middle term \(bx\) using the two numbers found.
4. Group the terms into two pairs.
5. Factor out the common factor from each pair.
6. Factor out the common binomial factor.

Example 2:

Factorise the trinomial:

2x² + 5x + 3

Solution:

1. Multiply the leading coefficient and the constant term: 2 × 3 = 6.
2. Find two numbers that multiply to 6 and add up to 5: 2 and 3.
3. Split the middle term using these numbers:
   2x² + 2x + 3x + 3
4. Group the terms into pairs:
   (2x² + 2x) + (3x + 3)
5. Factor out the common factor from each pair:
   2x(x + 1) + 3(x + 1)
6. Factor out the common binomial factor:
   (2x + 3)(x + 1)

6. Practice questions