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I think i have lost my brain

  1. Sep 4, 2005 #1
    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?
     
  2. jcsd
  3. Sep 4, 2005 #2

    LeonhardEuler

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    Gold Member

    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?
     
  4. Sep 4, 2005 #3
    its -6 and 2 isnt it?

    gosh i'm a fool sometimes :rofl:
     
  5. Sep 4, 2005 #4

    LeonhardEuler

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    That's it.
     
  6. Sep 4, 2005 #5
    arrrrrrgggghhhhhh

    how about x^2-6x-9
     
  7. Sep 4, 2005 #6

    LeonhardEuler

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    In this case the answer comes out irrational. Make sure you copied it right. Otherwise use the quadratic formula.
     
  8. Sep 5, 2005 #7

    HallsofIvy

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    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]
     
  9. Sep 6, 2005 #8
    I'm glad no one did the problem for him..... :rolleyes:
     
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