i think i have lost my brain

[aricho]

sorry this is such a stupid question.....but i have been sitting her for half an hour trying to solve it.

what does

a^2-4a-12

factorise to?

[LeonhardEuler]

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?

[aricho]

its -6 and 2 isnt it?

gosh i'm a fool sometimes :rofl:

[LeonhardEuler]

That's it.

[aricho]

arrrrrrgggghhhhhh

how about x^2-6x-9

[LeonhardEuler]

In this case the answer comes out irrational. Make sure you copied it right. Otherwise use the quadratic formula.

[HallsofIvy]

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 a2-4a-12= 0, a=
\frac{4\pm\sqrt{16-4(1)(-12)}}{2}= \frac{4\pm\sqrt{64}}{2}= \frac{4\pm8}{2} 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, x2-6x-9 does not have rational factors (are you sure this wasn't x2- 6x+ 9?) but can be factored if we solve the equation
x2- 6x+ 9= 0.

I would be inclined to "complete the square": x2- 6x= 9 and we complete the square on the left by adding (-6/2)2= (-3)2= 9 (no surprize there!). x2- 6x+ 9= (x- 3)2= 9+ 9= 18= 2*9.
x-3= \pm\3\sqrt{2}
so the solutions are x= 3- 3\sqrt{2} and x= 3+ 3\sqrt{2}
x^2-6x-9= (x-3+3\sqrt{2})(x-3-3\sqrt{2}).

You could also get that factorization by completing the square originally:
x^2- 6x- 9= x^2- 6x+ (9-9)- 9= x^2- 6x+ 9- 18= (x-3)^2- 18
Now factor as the difference of two squares:
x^2-6x-9= (x-3)^2-18= ((x-3)+\sqrt{18})((x-3)-\sqrt{18})
= (x-3+3\sqrt{2})(x-3-3\sqrt{3})

[sr6622]

I'm glad no one did the problem for him..... :rolleyes: