Reflection on Factorisation and Expansion

Reflection

Initially, I was confused by expansion and factorisation however after learning more about these two chapters, i have learnt more about it. I have also learnt how to apply it in my daily life. Also, I now know how to differentiate between expansion and factorisation. I am quite amazed at how math has so many ways to go about a question. It is something that I have learnt in these two chapters.

Factorisation by Grouping (Chap 4.4)

Factorisation of Algebraic Expressions by Grouping

Expressions containing four or more terms can be factorised by the method of grouping. In this method, the terms are divided into groups such that the terms in each group have a common factor. 

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

                  = (x+y)(a+b)

a) 3px - 9py - 4qx + 12 qy

         = 3p(x - 3y) - 4q(x - 3y)  

    (Extract common factor x+y)

=(x - 3y)(3p - 4q)

Factorisation Using Special Algebraic Identities (Chap 4.3)

Factorisation Using Special Algebraic Identities 

Since factorisation is the reverse of expansion, we have:

  a² + 2ab + b² = (a + b)²

a² - 2ab + b² = (a - b)²

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

a) 4x² - 25y² = (2x)² - (5y)² 

= (2x + 5y)(2x - 5y)  

 (apply a² - b² = (a + b)(a - b), where a = 2x and b= 5y)

Expansion Using Special Algebraic Identities (Chap 4.2)

Expansion using Special Algebraic Identities

Now we will learn how to expand special products of algebraic expressions: 

Perfect Squares and Difference of Two Squares

Perfect squares: 

(a + b)² = a² + 2ab + b² 

(a - b)² = a² - 2ab + b² 

Difference of two squares:

(a + b)(a - b) = a² - b² 

• Perfect Squares Example: 

a)  (x + 5)² = x² + 2(x)(5) + 5²    <------   Use (a + b)²  = a²  + 2ab +b² 

                   = x²  + 10x + 25                    Here a = x and b = 5.

b) (3x - 4)² = (3x)² - 2(3x)(4) + 4²   <------  Use (a - b)² = a²  - 2ab + b² 

               = 9x² - 24x + 16

*Be careful when expanding expressions like (x + 5)².

(x + 5)² ≠ x² + 25 

(x + 5)² = (x + 5)(x + 5)

• Difference of two Squares:

a)  (x+1)(x-1) = x² - 1²    <------ Use (a + b)(a - b) = a² - b²

                  = x² - 1 

b)  (x + y + 2)(x + y - 2)

    = [(x + y) + 2)][(x + y) - 2]    <------- Use (a+b)(a-b) = a² - b²

    = (x + y)² - 2²                                   Here a = (x + y) and b = 2

    = x² + 2xy + y² - 4     <------- Use (a+b)² = a² + 2ab + b².

Expansion and Factorisation of Algebraic Expressions (Chap 4.1)

Expansion and Factorisation of Algebraic Expressions

Example:

Expand and simplify each of the following expressions:

a) -3x (y+z) - 4x(2y-5z)

=-3xy - 3xz -8xy + 20xz (Distributive law)

=-3xy - 8xy -3xz + 20xz (Group like terms)

=-11xy + 17xz

b) 2x (x + 2y) + 3x(2x - 3y) = 2x² + 4xy + 6x² - 9xy (Distributive law)

= 2x² + 6x² + 4xy -9xy (Group like terms)

= 8x² -5xy

Factorisation of Quadratic Expressions (Chap 3.3)

Factorisation of Quadratic Expressions using a Multiplication Frame

a) Example: x² + 8x + 12

Step 1: Write x² in the top-left corner and 12 in the bottom-right corner of the multiplication frame.

Step 2: Consider the factors of x²  and 12. Write them in the first column and the first row.

Step 3: Multiply them to complete the multiplication frame and check whether the result matches the given expression.

Therefore, x²  - 5x + 4 = (x-1)(x-4)

b) Example: 3x²  + 7x -6 

Therefore, 3x²  + 7x - 6 = (3x - 2)(x + 3)

c) Example: 4x² - 6x - 4 = 2(2x² - 3x -2)  

extract the common factor 2

Therefore, 4x² - 6x - 4 = 2(2x + 1)(x - 2)

Expansion and Simplification of Quadratic Expressions (Chap 3.2)

Expansion and Simplification of Quadratic Expressions

a) Example: -2 (3x -2) 

    *Multiply 3x-2 by 2 before changing its sign. 

    Therefore, -2(3x-2) = (- 2)(3x)+(- 2)(- 2)

                                    = -6x + 4

b) Example: -2x(3x-2) = (-2x)(3x) + (-2x)(-2) 

                                     = -6x² + 4x

(Prezi link):

http://prezi.com/uf3po-ri-vba/expanding-and-simplifying-quadratic-equations/