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Factor Theorem

  1. Nov 22, 2009 #1
    I tried, doubt I'm even close to correct. Show me where I went wrong or just guide me with the problem please.

    1. The problem statement, all variables and given/known data
    1.) Prove that a + b is a factor of a²(b + c) + b²(c + a) + c²(a + b) + 2abc and write down the other two factors.

    2. The attempt at a solution

    [tex]a^2(b - a) + b^2(-a + a) + (-a)^2(a + b) + 2ab(-a) = 0[/tex]
    [tex]a^2b - a^3 - b^2a + b^2a + a^2 + ab - 2ab = 0[/tex]
    [tex]a^2b - a^3 + a^2 - ab = 0[/tex]


    [tex]a^2(-c + c) + (-c)^2(c +a) + c^2(a - c) + 2a(-c) = 0[/tex]
    [tex]-a^2c + a^2c + c^2 + ac + c^2a - c^3 - 2ac = 0[/tex]
    [tex]c^2 + c^2a - c^3 - ac = 0[/tex]

    [tex](-b)^2(b + c) + b^2(c - b) + c^2(-b + b) + 2(-b)bc = 0[/tex]
    [tex]b^2 + bc + b^2c - b^3 - c^2b + c^2b - 2bc = 0[/tex]
    [tex]b^2 + b^2c - b^3 - bc = 0[/tex]
     
  2. jcsd
  3. Nov 22, 2009 #2

    Mark44

    Staff: Mentor

    You're not given that a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc is equal to zero. All you're supposed to be doing is to rearrange what you have to put it into a factored form.

    Having said that, how does the following expression relate to the expression above?
    [tex]a^2(b - a) + b^2(-a + a) + (-a)^2(a + b) + 2ab(-a)[/tex]

    Also, are you sure that you have typed the problem exactly as it was given to you? I was able to factor the given expression into (b + c) times another factor, but I haven't been able to write it yet as (a + b) times another factor.
     
  4. Nov 22, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, a+ b is a factor. Think of this as a polynomial in a with b and c constants. a+ b= a-(-b) will be a factor if and only if setting a= -b makes the polynomial equal to 0. And, of course, you can find the other factor by dividing by a+b.
     
  5. Nov 22, 2009 #4
    Hi!!:smile:
    Solution to this is quite simple..
    1st open the brackets
    :. you have
    a2b+ab 2+b2c+c2b+a2c+c2a+2abc
    =ab(a+b) +b2c+a2c+c2b+c2a+2abc=ab(a+b)+c 2(a+b)+b2c+a2c+2abc
    = (ab+c 2)(a+b)+c(b 2+a 2+2ab)
    =(ab+c 2)(a+b)+c(a+b)2
    =(a+b)(c2+ab+ac+bc)
    That's what I think.
    (there could be some mistakes while typing in powers cuz I kinda get confused while typing them)
    I hope this helps!!:smile:
     
  6. Nov 22, 2009 #5
    Yes, I typed the question exactly as it was given.
    I'm not sure how to 'prove'. The teacher keeps giving these proofs and I get baffled by them.
     
  7. Nov 22, 2009 #6
    a + b = 0, and so
    a = -b or b = -a
     
    Last edited: Nov 22, 2009
  8. Nov 22, 2009 #7
    Hey whats the big deal then .
    You "PROVE" it by giving the definition of factor and as it suits this condition :. You can say it is a factor of it.
    See if it helps(?)
     
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