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Factoring a four term polynomial

  1. Feb 8, 2017 #1
    1. The problem statement, all variables and given/known data
    I just want to know how get from ##4x^3+3x^2-6x-5=0 ##
    to ##(x+1)^2(4x-5)=0##. What's the trick when dealing with these nasty polynomials? I got the answer by taking another approach (solving a root equation) but I noted this is also a way to go, but I can't figure out the method

    2. Relevant equations
    -



    3. The attempt at a solution
    Well I can't seem to get anywhere. The assignement is solved and correct, but I would like to learn how to make the factoring that out! =)


    Edit: corrected accidental + instead of - in the last factor.
    Edit 2: bad reception on smartphone didn't save the first correction. Now it should be fine.
     
    Last edited: Feb 8, 2017
  2. jcsd
  3. Feb 8, 2017 #2

    fresh_42

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    There must be some typos. Where do you get the negative signs from when multiplying ##(x+1)^2(4x+5)\,##?

    In general, you have basically two possibilities: you can look up the formulas to solve for ##4x^3 + \ldots = 0## which are a bit nasty, or, which is more promising in such designed cases, guess some roots. That is, if ##4a^3 + \ldots = 0## for some number ##a##, then ##(x-a) \,\vert \, (4x^3 + \ldots )## and you can perform a long division and solve for the remaining polynomial ##4x^2 + \ldots = 0##. To guess zeros, you can simply try low numbers as ##\{-2,-1,0,1,2\}## or you can look at the absolute term: ##4x^3 + \ldots - 5 = 4(x-a)(x-b)(x-c) = 0## shows that ##abc= 5## and you can try some divisors of ##5##.
     
  4. Feb 8, 2017 #3

    SammyS

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    Perhaps it should be ##\ 4x^3+3x^2-6x-5=0 \,.##
     
  5. Feb 8, 2017 #4

    epenguin

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    We get it, it's something like that except the opposite plus an accidental - instead of + in the second term.
     
  6. Feb 8, 2017 #5
    Yes, that's correct! (On mobile unit with bad reception). Edited it now! :)
     
  7. Feb 8, 2017 #6
    Yes! Two sign errors I'm afraid! Edited now. Oh okay, so if I guess a root and divide the polynomial with that factor (x-a) I'd be done? :) The factor theorem, right? (Still on mobile, gonna try this when I get home!)
     
  8. Feb 8, 2017 #7

    fresh_42

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    Yes. ##-1## as a root can easily be checked. Then you can calculate ##(4x^3+3x^2−6x−5):(x+1)## and get a quadratic polynomial for which you probably know the formulas to find the remaining zeroes (Vieta's formulas). Otherwise you can also repeat the process and in this case with ##-1## again. You have to divide by ##x+1 = x-(-1)## here, in general by ##x-a## if ##a## is a zero.
     
  9. Feb 8, 2017 #8

    Mark44

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    For your polynomial ##4x^3 + 3x^2 - 6x - 5##, the Rational Root Theorem can be used to determine whether there are factors of the form (x - p/q). This theorem says that any such roots p/q are such that p divides the constant (-5) and q divides the coefficient of the highest degree term (4). So the possibilities are ##\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4##. In all there are 12 possible rational roots. The easiest way to check each of them is by using Synthetic Division.

    Edit: Added my missing LaTeX end tag.
     
    Last edited: Feb 10, 2017
  10. Feb 8, 2017 #9

    ehild

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    You can do it in an intuitive way:smile:
    ##4x^3+3x^2-6x-5=3x^3+3x^2+x^3-x-5x-5=3x^2(x+1)+x(x-1)(x+1)-5(x+1)=(x+1)(4x^2-x-5)##
     
  11. Feb 8, 2017 #10

    epenguin

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    Still simpler than the methods suggested is simpe substitution of a likely looking root. In theexamples gi , -4 +3 -6 ven at your level roots are ften simple small integers. Look at x = 1 for instance the polynomial would be 4 + 3 - 6 - 5 ≠ 0, so not that, now try x = -1, -4 +3 +6 -5 = 0, yes -1 is a root, and (x + 1) is a factor, divide by (x + 1)...

    Also if the factors are simple rationals, I mean of form (Ax + B) with A, B integers, then the number of possibles to look at is limited, e.g. here the first term in that case must have come from
    (4x +...)(x +...)(x +...) or (2x +...)(2x +..)(x +...)
    and the last term from (... + 5)(.. + 1)(... -1) or (...- 5)(.. + 1)(... -1) or (... - 5)(.. - 1)(... -1),.

    (You only need tconsider the sign possibilities for one of these tw ,terms OK? :oldsmile: )
     
    Last edited: Feb 9, 2017
  12. Feb 10, 2017 #11

    haruspex

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    ##\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4##
     
  13. Feb 10, 2017 #12

    Mark44

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    Thanks! I didn't notice that I had neglected to add the closing tag. I have fixed it in my earlier post.
     
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