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How to factor these equations

  1. Feb 12, 2007 #1
    I dont know why but I have a lot of trouble factoring. After 3 years of it I would expect to be a lot better at it but I still find it as hard as when I first learned it. Im just wondering if anyone here can help me understand it a little different then my teachers have.

    To date I think I finally understand stuff like
    [tex]2x^2-2x-12[/tex]
    [tex]x^3-5x^2-6x[/tex]
    [tex]5x^3-45x[/tex]

    The stuff I am having trouble with is things like this
    [tex]x^3+x^2-4x-4[/tex]
    [tex]x^8-1[/tex]
    [tex]2m(m-n) + (m+n)(m-n)[/tex]

    Is there any certain rules or guidelines I may be able to follow to make these kind of questions easier. Right now im not really sure how to start them.
     
  2. jcsd
  3. Feb 12, 2007 #2

    cristo

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    For the first one if you can't factor it, then guess a factor and divide the polynomial by that factor to get a quadratic.

    The second one seems pretty straightfoward... what have you tried?

    In the third one, what is the common factor in both terms?
     
  4. Feb 12, 2007 #3
    The first one I understand now, I get (x+1)(x+2)(x-2)

    The second one im still not really sure

    Here is the third one I think
    (m-n)[2m + (m+n)]
    (3m+n)(m-n)
     
  5. Feb 12, 2007 #4
    I HATE factoring. I am great at all math I have ever tried(literally) and got a D on a factoring test. any thing beyond the complexity of x2 +ax +b i fail at. once x2 has a coefficient I fail miserably
     
  6. Feb 12, 2007 #5
    In order to understand the 3rd line better, Ill explain another way to factor the 1st line. You can group x^3 with x^2 and -4x with -4. Then you can factor each piece and youre left with:
    x^2(x+1) -4(x+1)
    This form is very similar to that third line. Next, you treat x^2 and -4 as coeffieients and factor out an (x+1) from both terms:
    (x+1)(x^2 - 4) This can then be reduced to what you got, and the third line also looks good

    But youre having the most trouble with the second line, x^8 - 1. Both terms are perfect squares. Whats the square root of x^8? Are you done after that, or can it be factored further?
     
  7. Feb 13, 2007 #6

    AlephZero

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    For problems where you know there is an easy answer (otherwise you wouldn't have been asked the question!) this often helps:

    If you want to factor [itex]ax^n \pm \dots \pm b[/itex] then for a factor of the form [itex]px \pm q[/itex] it is a good bet that p is a factor of a, and q is a factor of b.

    Also, use the remainder theorem: if (x-a) is a factor, the polynomial is zero when x = a. So in the [itex]x^3+x^2-4x-4[/itex] example it's fairly obvious the polynomial is zero when x = -1 therefore (x+1) is a factor.
     
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