Reflections

Reflections

Chapter 3:

I learnt new steps to solve the different algebra equations. Although I did not really understand these two chapters, I managed to understand the topic better by creating this blog and with the help of other online websites which have explanation of the topic. I managed to solve different equations and get a better grip of what I have learnt.

Chapter 4:

I have learnt the 3 different special algebraic identities and better understand how to factorise and expand the different equations. I have also learnt new things about the algebraic expressions and new rules/laws in algebra. Through this blog, I have learnt to understand better and I'm able to explain the things I have learnt carefully to everyone.

- Sarah Han (27) 2E4

Chapter 4

Further Expansion and Factorisation of Algebraic Expressions

The formula: a(b+c) = ab + ac is also know as the distributive law which we learnt in Chapter 3.

We can solve longer algebraic expressions with the formula too, but which more than one variable. For example:

(2x+4y)(3y+7x)

= 2x(3y+7x)+4y(3y+7x)

= 6xy + 14x² + 12y² + 28xy

=34xy +14x² + 12y²

Factorisation of Algebraic Expressions Using a 

Multiplication Table

I showed the multiplication table in Chapter 3, but I will show it again.

Like I have said previously, the the table can be to solve algebraic equations like x²+5+6x.

Expansion Using Special Algebraic Identities

The following 3 algebraic identities can help us solve algebraic equations easier.

1. (a+b)² 

= (a+b)(a+b) 

= a(a+b) + b(a+b)

= a² + ab + ab + b² 

= a² + 2ab + b²

2. (a-b)²

= (a-b)(a-b)

= a(a-b) - b(a-b)

= a² - ab - ab - b²

= a² - 2ab - b²

3.(a+b)(a-b)

= a(a-b) + b(a-b)

= a² - ab + ab - b²

= a² - b²

Let's try to solve an equation with one of these algebraic identities.

For example, let's solve this with the 1st algebraic identity:

(3x+6)²

= 3x² + 2(3x)(6) + 6²

= 3x² + 36x + 6²

= 3x² + 36 + 36

= 3x² + 72

Let's solve another one with the 2nd algebraic identity:

(5x-2y)²

= 5x² - 2(5x)(2y) - 2y²

= 5x² - 20xy - 2y²

Let's solve an example with the 3rd algebraic identity:

(3y+4)(3y-4) 

= (3y)² - (4)²

=9y² - 16

We can also solve problems like 120² with the use of special algebraic identities:

120²

=(100+20)²

=100² + 2(100)(20) + 20²

= 10000 + 4000 + 400

=14400

We can solve another problem with the algebraic identities too:

If (x + y)² = 361 and xy = -120, find the value of x² + y²

(x + y)² is the same as (a + b)²

so, (x + y)² = x² + 2xy + y² = 361

and if xy = -120,

x² + 2(-120) + y² = 361

x² + y² = 361 + 240

x² + y² = 601

Factorisation Using Specical Algebraic Identities

As we know the 3 different algebraic identities from above for expansion, factorising is just the opposite of it.
1. a² + 2ab + b² = (a + b)²
2. a² - 2ab + b² = (a - b)²
3. (a + b)(a - b) = a² - b²

Similarly, we can factorise some equations with these special algebraic identities too, and the 1st two algebraic identities can be solved by using the multiplication frame. For the 3rd identity:
4x² - 25y²
= (2x)² - (5y)²
= (2x - 5y)(2x + 5y)

We can also solve another problem using it:
103² - 9
= (100 - 3)² - 3²
= (103 + 3)(103 - 3)
=106 x 100
10600

Factorisation by grouping

Factorisation of Algebraic expressions of the form ax + ay. 

For example:
3x² + 9xy
=3x(x + 3y)

The common factor 3 and x have been taken out and factorising the equation.

Factorisation of Algebraic Expressions of the form ax + bx + kay + kby (k is the coefficient)

ax + bx + kay + kby = x(a + b) + ky(a + b)

= (a + b)(x + ky)

For example:

a(2x + 3) + 2(3 + 2x)

= (2x + 3)(a + 2)

Here's a video and you might understand more about quadratic equations especially factorisation!

Chapter 3

Expansion and Factorisation of Quadratic Equations

What is expansion?

Expansion is removing of the brackets of an equation by solving the equation inside the brackets first.

Let me show you a simple example:

2+(10-5)

= 2+5

=7

What is factorisation?

Factorisation is the opposite process of expansion, which is finding the common factor(s) to get an expression.

Let me show you a simple example:

5x + 5

= 5(x+1)                      5 is the common factor of 5x and 5.

Difference between expansion and factorisation

Expansion and factorisation is the opposite of one another. Expansion is to eliminate the brackets and to open up everything, while factorisation is to find the common factor and put it back into the brackets again.

Addition and Subtraction of Quadratic Expressions

x² = x X x

First off, let’s start with the simple addition and subtraction of quadratic equations

2x² + 4x² = 6x²

22x² – 10x² = 12x²

-4x-(-3x²) = =4x+3x²                   

*take note that x cannot be combined together because x and x^2 is two different terms.

Negative of a quadratic expression

For example:

-x+2 = -2-x 

This is because - is being multiplied to the both x and 2 inside the brackets.

Expansion and Simplification of Simple Quadratic Expressions

For example:

3(x+8) = 3x + 24

And another example:

and group the like terms together,

3a + 12 +4a +8 = 7a + 20

Expanding Quadratic Expressions of the form (a+b)(c+d)

For example, we have:

(3x+5)(2x+4) 

= 3x(2x+4) + 5(2x+4)

= 12x² + 12x + 10x + 20 For example:

= 12x² + 22x + 20

Another example:

Factorisation of Quadratic Expressions

For example:

x² + 5 + 6x = (x+1)(x+5)

we can form this table accordingly, and find the common factors of 5, and cross multiply both sides to form the right column. Add up the right column which will give the other term. The answer will be the equations on the left column, which is (x+1)(x+5).

This sums up the chapter!