To factorize a quadratic equation like this for example
X^2 + 2X - 12 = 0
after you've got the 2 X factors down do you just try out all the pairs of factors of the non variable number (in this case 12) and see if they work? For example

(X + ?) (X + ?)

the factors of 6 are:
12 x 1
6 x 2
3 x 4

so if this quadratic equation is true then 1 of these pairs of factors will always have to work? They don't appear to work when I plug them in. I can solve quadratics using the quadratic formula but this factorizing shit confuses me.

Some quadratics you cant factorise easily by just looking at them and you have to use the formula

Also if you don't know a way to avoid having to try every combination of numbers on those that can be done is to find a set of numbers that add to give the coefficient of x and times together to give the constant.

eg x^2 + 5x +6

numbers which add to give 5 and x to give 6 are 2 and 3 so they are what goes in the brackets

yes you have to use the quadratic formula. it is always possible youll get something like this, so you must use the another method.

if you use the quadratic formula, youll get

(-2 + sqrt(4 - 4*1*-12))/2 or (-2 - sqrt(4 - 4*1*-12))/2 which equals

-1 + sqrt(13) or -1 - sqrt(13)

and you wont be able an answer using the regular factorising method.

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there is also another method which is much more primitive, and is what the quadratic method is derived from. its called "completing the square. its longer, and you shouldnt use it. ill include it just for completeness
basically, its like this

X^2 + 2X - 12 = 0
you have to make the above a "perfect square". (ie, in the form (x - a)^2
we know that

X^2 + 2X + 1 = (X + 1)^2. but we have that -12 to deal with. so we just add 13 to both sides, so we get that +1 in the equation

X^2 + 2X - 12 + 13 = 0 + 13
X^2 + 2X + 1 = 13
(X + 1)^2 = 13 now we square root both sides
X + 1 = +sqrt(13) or -sqrt(13). now we subtract 1
X = -1 +sqrt(13) or -1 - sqrt(13) which is our answer
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EDIT 1

i would have to agree with gforce on this. if you look at the quadratic in the form x^2 + bx + c, 'b' will be the sum of the factors, and c will be the product. i would start by writing in the brackets (x___)(x___), then putting in the signs (eg. (x+__)(x-__)), then putting in the numbers

also, remember two things:
1) positive multiplied by positive or negative multiplied by negative is a positive number. conversly, positive multiplied by a negative is a negative number. this gives you clues. 'c' is negative, it must mean that the roots have different signs. if 'c' is positive, then they are either both negative or both positive.

2) if a positive number is subtracted by a positive number smaller than it, the difference will still be positive. conversely, if a positive number is subtraced by a positive number larger than it, the difference is negative. this may help you when finding roots. IF you have a negative 'c', ( which, but the point in (1), suggests that the roots are different in sign), if 'b' is positive, it means that the "larger" root will be the positive one, since youre subtracting a number by another number smaller than it. if 'b' is negative then the larger root is the negative one.

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EDIT 2:
ok, i would also like to inform you about what to do when the coefficient of the squared x is not 1. there are a few ways to do this, but this is how i do it.
eg.factorise
6x^2 + x - 2

ok lets write out our brackets first. however, this time, put the 6 inside every bracket, and divide the whole thing by 6. see below
((6x__)(6x__))/6

now, since the coefficient of x is just 1, the sum of the roots is 1. HOWEVER, the product of the root is the coefficient of x multiplied by c. that is, the product of the roots is 6*-2 = -12, NOT -2

'c' is a negative, which means the roots have different signs. 'b' is positive, so we know the larger root is the positive one. so
((6x +_)(6x -_)) /6
if the product is -12, and the sum is 1, we know the numbers must be 4 and -3

((6x + 4)(6x - 3)) /6
now, we get rid of the 6 in the denominator. we do this by first taking factorising the top
(2(3x + 2)3(2x - 1)) /6
then divide by 6
(6(3x + 2)(2x - 1)) /6

(3x + 2)(2x - 1) which is the factored form.

i find this to be faster than the quadratic methods sometimes, but do what you think is best.
remember THE QUADRATIC METHOD ALWAYS WORKS!(even with complex numbers, but i dont think youre up to that yet)

so heres a check list of what to do
1) does your quadratic have a constant? if there is no number 'c' such as X^2 + x, then just take out the x as a common factor. 0 is a root.
2)is your quadratic in the form ax^2 +2ab + b^2 ??? if so , its a PERFECT SQUARE (ax + b)^2
3)is your quadratic in the form x^2 - a^2 ?? if so, its a DIFFERENCE OF TWO SQUARES (x - a)(x + a)
4)is the coefficient of x^2 equal to 1? if so, factorise it by finding what the sum and products of the roots would equal to.
5) if that didnt work. use the quadratic formula x = (-b + sqrt(b^2 - 4ac))/2a or (-b - sqrt(b^2 - 4ac))/2a
6)if the coefficient of x^2 is more than 1 either use the method i did above, or the quadratic formula, which ever you find is faster.
7)if doing the first method, and it didnt work, use the quadratic formula

8)if you get a negative number in the square root, just right NO REAL ROOTS, since you cant have a negative root.

if you have done complex numbers before, there would be unreal roots, but i dont think your up to that yet

Thanks a lot Mantikore. That explains it all. I've memorized the quadratic formula and can factorize ones that can be factorized. I saw sites mention the complete the square method. I'm glad I don't have to learn that too.

Thanks a lot Mantikore. That explains it all. I've memorized the quadratic formula and can factorize ones that can be factorized. I saw sites mention the complete the square method. I'm glad I don't have to learn that too.

Oh but the complete the square method is oh so sexy once you start cranking the calculus.

Ummm... the quadratic formula IS completing the sqaure. Try completing the square for the generic quadratic polynomial ax^2 + bx +c, and you'll get the quadratic formula out of it.

this is how i learned it, nice and simple. i'll do this as an equation instead of just simplifying.
a- coefficient=1

ax^2+bx+c=0

1. x^2-6x+8=0 a=1 b=-6 c=8 ac=8
next simply find two numbers that multiply to ac (that being 8 in this example), as well as adding to b (that being -6)
so we now go through the multiples of ac (1)(8),(-1)(-8),(2)(4),(-2)(-4).
we can see that -2, and -4 both multiply to 8, but they also add up to -6. so now with these two numbers we break up the middle term (bx)

2. x^2-2x-4x+8=0 ---> you can see that -2x-4x=-6x so the equation has not changed
next place the two terms in parentheses like so
(x^2-2x)-(4x+8) now factor out what ever you can, but being sure that the upcoming two terms inside the parentheses are the same, as show here.
3. x(x-2)-4(x-2)] ---> you can now that the equation again has not changed, just expressed differently, x(x-2)=x^2-2x and
-4(x-2)=4x+8. we also notice the said "same terms" with in the parentheses, basically one is cancelled out. and now you have the simplified form
4. (x-4)(x-2)=0 since the equation equals zero one of these terms must equal 0 so solve the algebraic expressions
x-4=0 x-2=0
x=4 or x=2

NOW WITH A COEFFICIENT OTHER THAN 1
ax^2+bx+C
1. 9m^2-15m+4 a=9 b=-15 c=4 ac=36
again find what multiplies to ac and adds to b this is (-3)(-12)
break up the middle term
2. 9m^2-3m-12m=4
factor
3.3m(3m-1) -4(3m-1)
new terms are now
4.(3m-1)(3m+4) them solve

basically all you can do is just do a bunch of these. you'll eventually realize what a difference of two squares looks like, which is represented by (A-B)^2 which is when a is a perfect square, c is a perfect square and b ads up to half of c. looks like x-4x=4 (x-2)^2 and 4x^2-32x+64 (2x-8)^2 there is also a sum of squares which is (a+b)^2

these two types of squares are nice for harder equations that you are able to substitute in for a and b to avoid lots of writing and possible comfussion.