If you're ever absolutely stuck on a factorization problem, you can use the quadratic equation to find the roots. Then, suppose the roots are a and b. The equation will factor to (x-a)(x-b). In this case, that might not be necessary. Think: You need two numbers that multiply to give you -12 and add to give you -4. List the pairs of integers that you could multiply to get -12. Which of these add to give -4?
If you are every really, really stuck on a (quadratic) factoring problem you can do as Leonhard Euler said: solve the equation using the quadratic formula. If the roots are a and b, then the quadratic factors as (x-a)(x-b).
For example, to solve a^{2}-4a-12= 0, a=
[tex]\frac{4\pm\sqrt{16-4(1)(-12)}}{2}= \frac{4\pm\sqrt{64}}{2}= \frac{4\pm8}{2}[/tex] so that a= (4+8)/2= 6 and a= (4-8)/2= -2. The factors are (a-6)(a-(-2))= (a-6)(a+2).
More importantly, x^{2}-6x-9 does not have rational factors (are you sure this wasn't x^{2}- 6x+ 9?) but can be factored if we solve the equation
x^{2}- 6x+ 9= 0.
I would be inclined to "complete the square": x^{2}- 6x= 9 and we complete the square on the left by adding (-6/2)^{2}= (-3)^{2}= 9 (no surprize there!). x^{2}- 6x+ 9= (x- 3)^{2}= 9+ 9= 18= 2*9.
[tex]x-3= \pm\3\sqrt{2}[/tex]
so the solutions are [tex]x= 3- 3\sqrt{2}[/tex] and [tex]x= 3+ 3\sqrt{2}[/tex]
[tex]x^2-6x-9= (x-3+3\sqrt{2})(x-3-3\sqrt{2})[/tex].
You could also get that factorization by completing the square originally:
[tex]x^2- 6x- 9= x^2- 6x+ (9-9)- 9= x^2- 6x+ 9- 18= (x-3)^2- 18[/tex]
Now factor as the difference of two squares:
[tex]x^2-6x-9= (x-3)^2-18= ((x-3)+\sqrt{18})((x-3)-\sqrt{18})[/tex]
[tex]= (x-3+3\sqrt{2})(x-3-3\sqrt{3})[/tex]