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!