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Factoring problem

  1. Feb 7, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    I've tried the form (Aa+Bb+Cc)(Da+Eb+Fc) expand this and equate to the orig expression
    I ended up getting (5a+3b-5c)(6a+10b-6c) which when I expand does not give me a correct similar expression. I need help on this.
    Last edited: Feb 7, 2017
  2. jcsd
  3. Feb 7, 2017 #2


    Staff: Mentor

    This form can't be right. When you multiply this out you get terms to the 4th power. The expression you started with has terms only up to the 2nd power.
  4. Feb 7, 2017 #3

    Ray Vickson

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    This cannot possibly work: the product of two quadratic (2nd degree) expressions will be a 4th degree expression. However, you could try a product of two linear (1st degree) expressions---that would give you a 2nd degree expression, that might match the one you started with.
  5. Feb 7, 2017 #4
    What I mean is the form (Aa+Bb+Cc)(Da+Eb+Fc)

    Since Fc(Aa+Bb)+Cc(Da+Eb) = c(CE+BF) .........(1)


    where CFc^2=30 and (5a+3b)=(Aa+Bb), (6a+10b)=(Da+Eb), c(CE+BF)=-(61b+75a)c

    Now using (1)

    Fc(5a+3b)+Cc(6a+10b)=-(61b+75a)c chosing C=-5 and F=-6

    Therefore I have (5a+3b-5c)(6a+10b-6c) but the expansion of this is not (30a^2+68ab+30b^2)-61bc-75ac+30c^2

    I've tried this method to somehow find relation to the original expression but I still cant see it. Please help me.
  6. Feb 7, 2017 #5


    Staff: Mentor

    Is this the only candidate you tried? The coefficient of a2 is 30, but that doesn't necessarily mean that the factors have to be (5a + ...)(6a + ... ). Other pairs of numbers that multiply to 30 are 1 * 30, 2 * 15, and 3 * 10. Did you try any of these?
  7. Feb 7, 2017 #6
    Is there a way to avoid trial and error? I want a bullet-proof way of solving this.
  8. Feb 7, 2017 #7


    Staff: Mentor

    I don't think there's any way to avoid educated trial and error. By educated, I mean that I wouldn try 4 and 7.5 or any other combination in which both numbers aren't integers.
  9. Feb 7, 2017 #8
    I was thinking that maybe there's a method that I can use similar to factorization of the form (Aa+Bb+Cc)(Da+Eb+Fc)

    Fc(Aa+Bb)+Cc(Da+Eb) = c(CE+BF) the only trial and error here is to find factors C and F.

    Since the orig expression is of the form (Aa+Bb+Cc+Gd)(Da+Eb+Fc+Hd)
  10. Feb 11, 2017 #9
    You have AD = BE = CF = GH = 30, so you can eliminate 4 variables at once. D = 30/A etc.
    You then get 6 expressions like A (30/B) + B(30/A) = 68 by equating the coefficients of ab, ac, etc.
    You can find A/B from this equation.

    Note that a solution remains a solution if you multiply A,B,C,G with a constant r and divide D,E,F,H by R. You can just set A=1 to get a solution.Since you already know A/B the rest is easy.
    If you want integers for an answer, you might have to find a suitable r. (if you get B = 10/3 for example, just multiply A,B,C,G with 3 and divide D,E,F,H by 3.
  11. Feb 12, 2017 #10


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    You are making it unnecessarily complicated by considering so many variables at once. Start with just a and b, say. You have
    Clearly E=30/A, F=30/B. Writing x=A/B we get a quadratic equation for x.
  12. Feb 15, 2017 #11


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    The OP seems to have lost interest so I will lay out the approach in detail.

    Without loss of generality we may assume 1≤ A ≤ E. That means A is one of 1, 2, 3, 5. From looking at just the a and b terms, as above, we find the ratio between A and B is 3:5. So either A=3, B=5, E=10, F=6, or A=5, B=3, etc.
    Applying the same analysis to a and c we find the A to C ratio is -2 to 1, so C is -6 or -10.
    And so forth.

    It's not necessarily the fastest method, but it is fairly deterministic, only having to keep track of two possibilities.
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