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Math Help - factoring

  1. #1
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    factoring

    how do i factor 0= x^5-x^4+14x^3-16x^2-32x?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: factoring

    Quote Originally Posted by cheshirecat View Post
    how do i factor 0= x^5-x^4+14x^3-16x^2-32x?


    x^5-x^4+14x^3-16x^2-32x=

    x(x^4-x^3+14x^2-16x-32)=

    one root is obviously x_1=0, the second is one from the group : {1,-1,2,-2,4,-4,8,-8,16,-16,32,-32}. It's easy to see that x_2=2 and x_3=-1 are roots.

    Now, divide x^4-x^3+14x^2-16x-32 by (x-2)(x+1), what would you get?
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  3. #3
    Super Member TheChaz's Avatar
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    Re: factoring

    Quote Originally Posted by cheshirecat View Post
    how do i factor 0= x^5-x^4+14x^3-16x^2-32x?
    Theory suggests using the "rational root theorem" to narrow down the possible roots to those that AsZ gave.
    Then "synthetic division" will make it rather painless to determine if a given contender is indeed a root.
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  4. #4
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    Re: factoring

    Quote Originally Posted by Also sprach Zarathustra View Post
    x^5-x^4+14x^3-16x^2-32x=

    x(x^4-x^3+14x^2-16x-32)=

    one root is obviously x_1=0, the second is one from the group : {1,-1,2,-2,4,-4,8,-8,16,-16,32,-32}. It's easy to see that x_2=2 and x_3=-1 are roots.

    Now, divide x^4-x^3+14x^2-16x-32 by (x-2)(x+1), what would you get?
    I don't know how to divide (x-2)(x+1) into x^4-x^3+14x^2-16x-32. I know how to get the 3 real roots by using synthetic division which gives x(x+1)(x-2), but I cannot figure out how to get the (x^2 + 16) part of the answer. I know that gives the 2 complex solutions 4i and -4i and I also don't know how they knew there were 2 complex solutions.
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  5. #5
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    Re: factoring

    Code:
    -1]...1...-1...14...-16...-32
    ..........-1....2...-16....32
    -----------------------------
    ......1...-2...16...-32.....0
    
    2]....1...-2...16...-32
    ...........2....0....32
    -----------------------
    ......1....0...16.....0
    note the coefficients of the final depressed polynomial ...

    x^2 + 16
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  6. #6
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    Re: factoring

    Quote Originally Posted by skeeter View Post
    Code:
    -1]...1...-1...14...-16...-32
    ..........-1....2...-16....32
    -----------------------------
    ......1...-2...16...-32.....0
    
    2]....1...-2...16...-32
    ...........2....0....32
    -----------------------
    ......1....0...16.....0
    note the coefficients of the final depressed polynomial ...

    x^2 + 16
    But how did you know to pick those two coefficients from that particular root? How did you know using synthetic division there were 2 complex roots?
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  7. #7
    Super Member Quacky's Avatar
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    Re: factoring

    How I would approach: use an appropriate theorem to find the first three roots.

    Then, we have:

    0=x^5-x^4+14x^3-16x^2-32x

    =x(x+1)(x-2)(ax^2+bx+c)

    =x(x^2-x-2)(ax^2+bx+c)

    we can then equate coefficients.

    In order to have 1x^5, we can quickly see that a=1

    =x(x^2-x-2)(x^2+bx+c)

    =(x^3-x^2-2x)(x^2+bx+c)

    Now we can look at the x^4 terms. If we expand this, but only by looking at the terms that will give us an x^4, we get:

    x^3(bx)-x^2(x^2)=-x^4

    bx^4-x^4=-x^4

    b=0

    Then, looking at the terms involving just x:

    -2c=-32

    c=16

    Thus making the factors:

    0=x(x+1)(x-2)(x^2+16)

    At this point, does common sense not dictate that x^2+16\neq{0} for real x?
    Last edited by Quacky; November 18th 2011 at 09:47 PM.
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  8. #8
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    Re: factoring

    Quote Originally Posted by cheshirecat View Post
    But how did you know to pick those two coefficients from that particular root? How did you know using synthetic division there were 2 complex roots?
    how do u do this part?
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  9. #9
    Super Member TheChaz's Avatar
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    Re: factoring

    You just GUESS. We determined that there are only a handful of potential rational roots. I like to start small, with +/-1. Then move onto the larger ones. It takes two successful trials to reduce a 4th degree polynomial to a quadratic, at which point you can use the quadratic formula.
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