Factoring Quadratics
To "Factor" (or "Factorise" in the UK) a Quadratic is to:
find what to multiply to get the Quadratic
It is called "Factoring" because you find the factors (a factor is something you multiply by)
Example
The factors of x^{2} + 3x  4 are:
(x+4) and (x1)
You can see this by simply expanding (x+4)(x1):
(x+4)(x1) 
= x^{2}  x + 4x  4 

= x^{2} + 3x  4 
(See Expanding and Multiplying Polynomials if you don't understand what I did there)
In fact, Expanding and Factoring are opposites:
Expanding is easy, but Factoring can often be tricky

It can be like trying to find out what ingredients went into a cake to make it so delicious.
It is sometimes not obvious at all! 
So let us try an example where we don't know the factors yet:
Example: what are the factors of 2x^{2} + 7x + 3 ?
We could guess (2x+3)(x+1):
(2x+3)(x+1) = 2x^{2} + 2x + 3x + 3 = 2x^{2} + 5x + 3 (WRONG)
How about (2x+7)(x1):
(2x+7)(x1) = 2x^{2}  2x + 7x  7 = 2x^{2} + 5x  7 (WRONG AGAIN)
OK, how about (2x+9)(x1):
(2x+9)(x1) = 2x^{2}  2x + 9x  9 = 2x^{2} + 7x  9 (WRONG AGAIN)
Oh No! We could be guessing for a long time before we get lucky!
A Method For Simple Cases
Luckily there is a method that works in simple cases.
Step 1: Find two numbers that add to give b, and multiply to give ac
Example: 2x^{2} + 7x + 3
b is 7, and ac is 2×3 = 6
So we want two numbers that add up to 7, and multiply together to make 6.
In fact 6 and 1 do that (6+1=7, and 6×1=6)
Step 2: Rewrite the middle with those numbers:
Rewrite 7x with 6 and 1:
2x^{2} + 6x + x + 3
Step 3: Factor the first two and last two terms separately:
The first two terms 2x^{2} + 6x factor into 2x(x+3)
The last two terms x+3 don't actually change in this case
So we get:
2x(x+3) + (x+3)
Step 4: If you've done this correctly, your two new terms should have a clearly visible common factor.
In this case you can see that (x+3) is common to both terms
So we can now rewrite it like this:
2x(x+3) + (x+3) = (2x+1)(x+3)
Check: (2x+1)(x+3) = 2x^{2} + 6x + x + 3 = 2x^{2} + 7x + 3 (Yes)
Much better than guessing!
Let us try another example:
Example: 6x^{2} + 5x  6
Step 1: b is 5, and ac is 6×(6) = 36
One of the numbers has to be negative to make 36, so by playing with a few different numbers I find that 4 and 9 work nicely:
4+9 = 5, and 4×9 = 36
Step 2: Rewrite 5x with 4 and 9:
6x^{2}  4x + 9x  6
Step 3: Factor first two and last two:
2x(3x  2) + 3(3x 2)
Step 4: Common Factor is (3x  2):
(2x+3)(3x  2)
Check: (2x+3)(3x  2) = 6x^{2}  4x + 9x  6 = 6x^{2} + 5x  6 (Yes)
Why Factor?
Well, one of the big benefits of factoring is that we can find the roots of the quadratic equation (where the equation is zero).
All we need to do (after factoring) is find where each of the two factors becomes zero
Example: what are the roots of 6x^{2} + 5x  6 = 0 ?
We already know the factors are
(2x + 3)(3x  2)
And we can figure out that
(2x + 3) would be zero when x = 3/2
and
(3x  2) would be zero when x = 2/3
So the roots of 6x^{2} + 5x  6 are:
3/2 and 2/3
Check (a bit of arithmetic!):
3/2: 6(3/2)^{2} + 5(3/2)  6 = 6×(9/4)  15/2  6 = 54/4  15/2  6 = 66 = 0
2/3: 6(2/3)^{2} + 5(2/3)  6 = 6×(4/9) + 10/3  6 = 36/9 + 10/3  6 = 66 = 0
The General Solution
There is also a general solution (useful when the above method fails), which uses the quadratic formula:
If you use that formula and get the two answers x_{+} and x_{} (one is for the "+" case, and the other is for the "" case in the "±"), you can get this factoring:
a(x  x_{+})(x  x_{})
Let us use the previous example to show you how this one works:
Example: what are the roots of 6x^{2} + 5x  6 = 0 ?
Substitute a=6, b=5 and c=6 into the formula:
x = (b ± √[b^{2}  4ac]) / 2a
x = (5 ± √[5^{2}  4·6·(6)]) / 2·6
= (5 ± √[25^{} + 144]) / 12
= (5 ± √169) / 12
= (5 ± 13) / 12
So the two roots are:
x_{+} = (5 + 13) / 12 = 8/12 = 2/3,
x_{} = (5  13) / 12 = 18/12 = 3/2
(Notice that we get the same result we did with the factoring we used before)
Now put those values into a(x + x_{+})(x  x_{}):
6(x  2/3)(x + 3/2)
We can rearrange that a little to simplify it:
3(x  2/3)·2(x + 3/2) = (3x  2)(2x + 3)
And we get the same factors as we did before!
(Thanks to "mathsyperson" for parts of this article)
