((x²-x-12))/x)/(x/(x-4))


Okay, so let's start piece by piece and see where we go.

First of all, you can factor (x²-x-12) to (x-4)(x+3).  This gives ((x-4)(x+3))/x)/(x/(x-4)).

Now, remember how A/B is the same as A times (1/B)?  Well, the same thing works with A/(B/C), however, 1/(B/C) is the same as C/B.  So A/(B/C) is the same as A times (C/B).  This means that ((x-4)(x+3))/x) which is A in this case, divided by (x/(x-4)) which is B/C in this case, gives ((x-4)(x+3))/x) times ((x-4)/x).  Combining these gives ((x-4)(x-4)(x+3)/x²).

Now, I don't know exactly where you want to go with this, but that is pretty simplified there.  You could multiply the numerator terms together and get (x²-x-12)(x-4) or (x³-5x²-16x+48).

So (x³-5x²-16x+48)/x² is an answer.

EDIT: Also, if you set it equal to zero, you could disregard that x² and use (x-4)(x-4)(x+3)=0.  So you have x=4, 4, and -3.

I'm also learning polynomial factorization at school now, and tried it too, but why can you do this:

First of all, you can factor (x^2-x-12) to (x-4)(x+3).  This gives ((x-4)(x+3))/x)/(x/(x-4)).

I have no idea.

First of all, the A, B, and C are different from my post above.  The general form of a second degree polynomial (ie, one with it's highest term being x²) is Ax²+Bx+C.  In the case above, A=1, B=-1, C=-12.

Now, to multiply something of the form (d*x+p)(e*x+q), we use a process called FOIL.  This stands for First, Outer, Inner, Last.  This means you first multiply the first terms together, which is d*x and e*x in this case, to get d*e*x².  The Outer refers to the first term, and the last term being multiplied together.  In this case, d*x and q are multiplied together to get e*x*q.  Inner refers to the two inside terms, e*x and p. Multiplying them together gives px.  Also note that p and q are usually constants, so you can combine them by (e*p+d*q)x.  Lastly, or Last as it were, we have p and q being multiplied together, to get pq.

Now, remember the A, B, and C in the Ax²+Bx+C.  To find the d, e, p, and q, that fits a certain A, B, and C, we have to figure these equations out:
A=d*e
B=e*p+d*q
C=p*q
Now, A is 1 in this case, so it simplifies to:
A=1
B=p+q
C=p*q
B is -1 and C is -12.  In this case, -4 and 3 work because -3*4=-12=C and -3+4=-1=B.

Does that all make sense?

Isnt there a formula to factor second degree equations of the type: ax^2+bx+c. I think there is, what is it?

Yes, of course, the quadratic formula.  Let me link to wolframalpha

***BUMP***

How to factorize:

ax^2+bx+c

Using a formula, not the hand method.

Any ideas? Thanks.

If you find the solutions, ie, x=4 and x=-3, then it is simply x-4=0 and x+3=0 or (x+3)(x-4)=0.  That's only if A=1 though.  I don't think there is a formula, considering the complexity is basically that of a quadratic formula.  There are methods like completing the square that help, though.

Weird, my teacher gave us a formula, if only I had copied it to my notebook as he asked...

you mean stuffs like this?

Difference of Squares = (A²-B²) = (A-B)(A+B)
Sum of Squares = (A²+B²) = (A-Bi)(A+Bi)

Nah, those are the 'notable cases', special cases.