Short description:
Factorisation is the decomposition of an expression into multiple products of ‘simpler’ terms. It has a wide range of uses. For example, it can be used for solving polynomial equations, proving irrationality, drawing graphs, and so on.


How it works:
1) 24 = 6x4
Here we have decomposed 24 into the product of 6 and 4. Note
the factorisation may not be unique. For example, 24 = 12x2 = 2x2x2x3. (In fact we can factorise any integer into just multiples of primes, called its prime factorisation)

It also works for more general cases where x takes any real
number:
2) 3x^2+3 = 3(x^2+1)
3) 3x^2+12x = 3x(x+4)
4) 3x^2+4x+1 = (3x+1)(x+1)


You can check whether this works or not by multiplying the right hand side out to see if you get the left hand size expression (just as you would for (1) where 4x6 is indeed 24)

Expressions of the form like (4) (i.e. quadratic expressions that look like this: ax^2+bx+c where
a,b,c are constants) may be difficult to factorise.
Here is a way to approach it:
When given something like ax^2+bx+c consider all the factors of a and c.
And then try and add each factor of a to each factor of c to get b.

E.g. consider 3x^2+x-2
So a = 3, b=1, c=-2
Factors of a are 1,3,-1,-3 (2 pairs of factors: 1x3 and -1x-3)
Factors of c are -1,2,1,-2 (2 pairs of factors: -1x2 and 1x-2)
So try and pick one factor of a and one factor of c to add to 1 (because b=1)

3+(-2)=1 so 3 and -2 work.

Now take 3 and -2 along with each of its other factor of a so we have 1,3,-2,1.
And finally we place the four numbers into the slots of this (_x+_)(_x+_)
The answer is ((3)x+(-2))((1)x+(1)) = (3x-2)(x+1)



Examples:
1) Solve the equation 3x^2-6x+3 = 0
2) Solve the equation 5x^2+7x-6 = 0

Solution:
1) 3x^2-6x+3 = 3(x^2-2x+1) = 0
So x^2-2x+1 = 0
x^2-2x+1 = (x-1)(x-1) = 0
So x-1 = 0 which means x = 0
2) 5x^2+7x-6 = (5x-3)(x+2) = 0
So either (5x-3) = 0 which means x = 3/5
Or (x+2) = 0 which means x = -2
And so the solution is x = 3/5 or -2