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Factoring problem - is it an iteration or more obvious?

  1. Apr 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A part of a problem is to factor 8 - b^3

    "8 minus b cubed"

    2. Relevant equations



    3. The attempt at a solution

    I see that the problem is 2^3 - b^3.

    I don't see the next step

    I know the answer is (2-b)(4+2b+b^2).

    but I can't get there.

    Is this just trail and error dividing terms into the polynomial or is there a more obvious solution?

    Thanks
    Sparky_
     
  2. jcsd
  3. Apr 26, 2008 #2

    dynamicsolo

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    Are you familiar with polynomial division? (I'm assuming you want an answer better than 'use the difference of two cubes formula'.)

    You would look for a value of b which solves the equation
    8 - (b^3) = 0 , which gives b^3 = 8 and thus b = 2. The factors of a polynomials can be expressed as differences of the variable and the zeroes of the polynomials (are you familiar with this theorem?). So we now know that one factor of 8 - (b^3) is
    (b - 2) . If you divide this factor into the original polynomial, you will get - ( 4 + 2b + b^2 ).
     
    Last edited: Apr 26, 2008
  4. Apr 27, 2008 #3

    HallsofIvy

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    (Intelligent) trial and error is pretty much the way you (try to) factor most polynomials (I say "try to" because, of course, most polynomials can't be factored using only integer coefficients).

    The other way is to memorize some basic formulas. In particular, [itex]a^n- b^n= (a- b)(a^{n-1}+ a^{n-2}b+ a^{n-3}b^2+ \cdot\cdot\cdot+ a^2b^{n-3}+ ab^{n-2}+ b^{n-1})[/itex] is useful for this problem. It is also true that if n is odd then [itex]a^n+ b^n= (a+b)(a^{n-1}- ba^{n-2}+ b^2a^{n-3}- \cdot\cdot\cdot+a^2b^{n-3}- ab^{n-2}+ b^{n-1})[/itex].

    Of course, if you don't require that the coefficients be integer, there is a sure-fire method of factoring any polynomial, p(x). First find all roots of the equation p(x)= 0, say [itex]x_1, x_2, \cdot\cdot\cdot , x_n[/itex] where n is the degree of the polynomial, including complex roots and counting the correct multiplicity for each root. Then, if the leading coefficient is a, [itex]p(x)= (x- x_1)(x- x_2)\cdot\cdot\cdot(x- x_{n-1})(x-x_n)[/itex]. Of course, that method is not terribly useful if you want to factor in order to solve the equation!
     
  5. Apr 27, 2008 #4

    tiny-tim

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    Hi Sparky! :smile:

    Substitute b = 2a.

    Then you have 8(1 - a³).

    Do you know how to factor 1 - a³ ? :smile:
     
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