UPSB v3

Off-topic / Math halp

  1. Mike
    Date: Thu, Oct 29 2009 22:29:16

    Are you able to factor this?

    x^3+x-3

  2. KunLin
    Date: Fri, Oct 30 2009 03:19:49

    x is not a real number

  3. Awesome
    Date: Fri, Oct 30 2009 03:36:07

    QUOTE (KunLin @ Oct 29 2009, 11:19 PM) <{POST_SNAPBACK}>
    x is not a real number

    what?

    x^3+x-3

    I would say no but you could go

    x(x^2+1) -3

    but that kinda eliminates the point of factoring and is not solving by factoring even.

  4. Dark Angel-REX
    Date: Fri, Oct 30 2009 05:24:59

    QUOTE (Mike @ Oct 30 2009, 07:29 AM) <{POST_SNAPBACK}>
    Are you able to factor this?

    x^3+x-3


    Let
    f(x) = x^3 + x - 3

    when
    f(x) = 0

    x^3 + x - 3 = 0

    when you solve for x, you'll get complex number. You can still factor it, but solving this in high-school is impossible.

  5. catfish
    Date: Fri, Oct 30 2009 08:42:58

    i dont think you can factor that

  6. Dark Angel-REX
    Date: Fri, Oct 30 2009 12:51:32

    you can factor it. You definitely can. As long as you know the solution to x when the f(x)=0, you can. say that the solution to f(x)=0 is a, b, and c. (a, b, and c can be imaginary number too).
    so you can factor it like this (x - a)(x - b )(x - c)

    There are 3 solution to f(x) = 0. 2 of them are imaginary number. one of them is complex number.
    You wouldn't find out the answer. It'll take forever. try using online mathematics solution site.

  7. Tialys
    Date: Fri, Oct 30 2009 18:24:09

    You can't factor it exactly, only approximately, since the root is irrational. Using Newton's iterative procedure (Newton-Raphson method), you can find a real root. In this case, x ~ 1.21341. Then you can factor it out:

    x3 + x - 3
    = (x - 1.21341)(x2 + 1.21341x + 3/1.21341)
    = (x - 1.21341)(x2 + 1.21341x + 2.47237)

    The quadratic part can be factored further using the quadratic equation:

    ax2 + bx + c
    x = [-b +/- (b2 - 4ac)1/2]/2a

    The value of the discriminant (part under the square root: b2 - 4ac) will tell you if the remaining two roots are real or complex. For this function, (1.21341)2 - 4(1)(2.47237) < 0, so these roots are complex.

  8. Dark Angel-REX
    Date: Sat, Oct 31 2009 13:52:35

    you can factor it even if the solutions are irrational or even imaginary.