Is there any way to factor this x^3 - 5x^2 - 8x + 4 ???

Discussion in 'General Math' started by Eddie, Apr 16, 2007.

  1. Eddie

    Eddie Guest

    Hi,

    I need to factor [ x^3 - 5x^2 - 8x + 4 ] in order to solve a
    recurrence for an algorithms class, but I seem to be stuck. The
    closest I've been able to get is (x - 1)(x - 2)(x - 2), which yields
    [ x^3 - 5x^2 + 8x - 4 ] , but as you can see, the signs on the 3rd and
    last term are off.

    The particular method for solving the recurrence strictly calls for
    factoring of this polynomial, but I'm starting to think this
    polynomial can't be factored. Anyway, before I say this to my
    professor, I want to make sure there is no other way to factor
    this....

    Please help! (Thanks).

    Ed
     
    Eddie, Apr 16, 2007
    #1
    1. Advertisements

  2. Eddie

    Felicis Guest

    Must it be factored into three monomials with integer coefficients?
     
    Felicis, Apr 16, 2007
    #2
    1. Advertisements

  3. Hint: If the roots are r1, r2, and r3, what is their product? What
    does this say about
    possible rational roots?
    Hint: If the polynomial is reducible then it must have at least one
    linear factor and correspondingly
    at least one rational root.
    Hint: Since the polynomial is monic, if it has a rational root, then
    one can conclude
    something further about that root.
    Hint: If you find one root, (say) r, then you can divide by (x-r)
    to get a quadratic.
    I assume you know how to factor a quadratic.
     
    Pubkeybreaker, Apr 16, 2007
    #3
  4. Eddie

    Eddie Guest

    Hmm... I belive so because I need to plug-in both the roots and the
    coefficients into the general solution for the recurrence I'm trying
    to solve (which is of the form [ a_n = C_1r_1^n + C_2r_2^n +
    C_3r_3^n ] (note I used the underscore to indicate a subscript).
     
    Eddie, Apr 16, 2007
    #4
  5. Eddie

    Temp Guest

    No "nice" roots. There are a number of online polynomial root
    finders. Here's one:
    <http://home.att.net/~srschmitt/script_cubic.html>
     
    Temp, Apr 16, 2007
    #5
  6. Eddie

    Mate Guest


    (x-14*COS(2*ATAN(SQRT(1518)/99)/3)/3-5/3)*
    (x+14*SIN(2*ATAN(SQRT(1518)/99)/3+pi/6)/3-5/3)*
    (x+14*COS(2*ACOT(-SQRT(1518)/99)/3)/3-5/3)


    Mate
     
    Mate, Apr 16, 2007
    #6
  7. Eddie

    W. Dale Hall Guest

    Are you sure you have the signs correct? If this were

    x^3 - 5 x^2 + 8 x - 4

    then it would factor easily.

    Dale
     
    W. Dale Hall, Apr 17, 2007
    #7
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.