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I think i have lost my brain

  • Thread starter aricho
  • Start date
71
0
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?
 

Answers and Replies

LeonhardEuler
Gold Member
858
1
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?
 
71
0
its -6 and 2 isnt it?

gosh i'm a fool sometimes :rofl:
 
LeonhardEuler
Gold Member
858
1
That's it.
 
71
0
arrrrrrgggghhhhhh

how about x^2-6x-9
 
LeonhardEuler
Gold Member
858
1
In this case the answer comes out irrational. Make sure you copied it right. Otherwise use the quadratic formula.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
898
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=
[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, 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.
[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]
 
36
0
I'm glad no one did the problem for him..... :rolleyes:
 

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