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Math Help - Factoring Polynomials

  1. #1
    Newbie nickgc's Avatar
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    Factoring Polynomials

    Hi forum!

    I was factoring this polynomial and i got stucked, please help me..

    x^5 - 6x^4 + 5x^2 - 16x^2 - 12x - 16

    what i did is i grouped them:

    ( x^5 - 6x^4 + 5x^2 ) + ( - 16x^2 - 12x - 16 )

    Then i factored each group:

    (x^3) ( x - 5 ) ( x - 1 ) + [ -4 ( 4x^2 - 3x - 4 )]

    then i don't know what to do next...

    i have tried doing trials and errors using synthetic division but i can't solve it.. help me please..

    thanks forum
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  2. #2
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    Re: Factoring Polynomials

    It doesn't factorise.
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  3. #3
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    Re: Factoring Polynomials

    ok thanks!

    but how would i know if a polynomial doesn't factorise?
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  4. #4
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    Re: Factoring Polynomials

    Use a CAS...
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  5. #5
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    Re: Factoring Polynomials

    I'm sorry but i am not familiar with CAS, can you help me with it?
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    Re: Factoring Polynomials

    Quote Originally Posted by nickgc View Post
    I'm sorry but i am not familiar with CAS, can you help me with it?
    A Computer Algebra System - basically, use a calculator if you can't see a simple factor.
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  7. #7
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    Re: Factoring Polynomials

    thank you
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    Re: Factoring Polynomials

    Quote Originally Posted by nickgc View Post
    Hi forum!

    I was factoring this polynomial and i got stucked, please help me..

    x^5 - 6x^4 + 5x^2 - 16x^2 - 12x - 16

    [snip]
    Why are you trying to factorise it?
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  9. #9
    Newbie nickgc's Avatar
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    Re: Factoring Polynomials

    I am reviewing for our midterm exams...
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    Re: Factoring Polynomials

    Quote Originally Posted by Prove It View Post
    It doesn't factorise.
    The fundamental theorem of algebra says it does. But it has no rational roots though it has one real root near 5.696.

    CB
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    Re: Factoring Polynomials

    Quote Originally Posted by nickgc View Post
    I am reviewing for our midterm exams...
    That does not explain why you're trying to factorise it!

    What I was hoping you would explain in repsonse to my question was - Did the question say or suggest you had the factorise it?
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  12. #12
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    Re: Factoring Polynomials

    I'm sorry

    Yes, the instruction was to factorise it
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    Re: Factoring Polynomials

    Quote Originally Posted by CaptainBlack View Post
    The fundamental theorem of algebra says it does. But it has no rational roots though it has one real root near 5.696.

    CB
    Which means that it does not have any "nice" linear factors, but that does not mean that it does not factor into the product of a "nice" quadratic and a "nice" cubic. Though in this case I think not.

    CB
    Last edited by CaptainBlack; July 18th 2011 at 06:52 AM.
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    Re: Factoring Polynomials

    Quote Originally Posted by nickgc View Post
    Hi forum!

    I was factoring this polynomial and i got stucked, please help me..

    x^5 - 6x^4 + 5x^2 - 16x^2 - 12x - 16
    Assuming the third term ought to be 5x^3

    if you had a typo on the 4th term also, (+16 instead of -16)
    you would have a polynomial which has 3 factors (some repeated)
    and 3 different integer roots for f(x) = 0.
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  15. #15
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    Re: Factoring Polynomials

    Assume that f=x^5-6x^4-11x^2-12x-16 can decompose into rational polynomials it will be of form (ax^2+bx+c)(dx^2+ex+f)(x-q),a,b,c,d,e,f,q\in Q so it will have a rational root assume \frac{p}{q},(p,q)=1so (\frac{p}{q})^5-6(\frac{p}{q})^4-11(\frac{p}{q})^2-12\frac{p}{q}-16=0 \Rightarrow p^5-6p^4q-11p^2q^3-12pq^4-16q^5=0 \Rightarrow p|q,q|p\Rightarrow p=q contradiction.
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