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Math Help - roots

  1. #1
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    roots

    may i know what method should i use to find the roots of 3x^5 - 7x^4 + 11x^3 = 21x^2 - 22x + 8? can anybody get my confusion out from head please.

    thank you
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    Re: roots

    Rearrange to get: 3x^5 - 7x^4+11x^3-21x^2+22x-8 = 0. By inspection it's clear that one root is x= 1. So divide through by (x-1) and you'll get a polynomial of 4th order, which it also turns out has a root of x=1. Continue until you get it to a 2nd order whose roots can be found "the usual way" using the quadratic equation.
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    Re: roots

    Wolfram Alpha rearrangement and complex roots (if you need them)

    You can solve quintics by factoring them into smaller parts (permutations or radicals). See Quintic Equation -- from Wolfram MathWorld.
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    Re: roots

    thanks you bopbop ... can you show the solution bopbop?

    thanks
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    Re: roots

    i was using the synthetic division on this by x - 1... then there is remainder after i solved it in first value of divisor to be 1... meaning there is no roots but root ( 1 root ) . second question, is it considered a root when there is remainder of 16/ x-1?
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  6. #6
    MHF Contributor ebaines's Avatar
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    Re: roots

    If you do the division of the polynomial by (x-1) three times you get:

    3x^5 - 7x^4+11x^3-21x^2+22x-8 = (x-1)(3x^4-4x^3+7x^2-14x+8) = (x-1)^2(3x^3-x^2+6x-1) = (x-1)^3(3x^2+2x+1)

    So the roots are 1 and the two solutions to 3x^2+2x+1=0.
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    Re: roots

    Quote Originally Posted by ebaines View Post
    If you do the division of the polynomial by (x-1) three times you get:

    3x^5 - 7x^4+11x^3-21x^2+22x-8 = (x-1)(3x^4-4x^3+7x^2-14x+8) = (x-1)^2(3x^3-x^2+6x-1) = (x-1)^3(3x^2+2x+1)

    So the roots are 1 and the two solutions to 3x^2+2x+1=0.
    You lost an 8 in there. The answer should be (x-1)^3(3x^2+2x+8).

    (3x^2+2x+8) is irreducible.
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    MHF Contributor ebaines's Avatar
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    Re: roots

    Quote Originally Posted by Bean View Post
    You lost an 8 in there. The answer should be (x-1)^3(3x^2+2x+8).
    Right - typo on my part, good catch.

    Quote Originally Posted by Bean View Post
    (3x^2+2x+8) is irreducible.
    Apply the quadratic formula, and you find 3x^2 + 2x + 8 = 3(x + \frac 1 3 + \frac 2 3 \sqrt{23} i  )(x + \frac 1 3 - \frac 2 3 \sqrt{23} i  )
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  9. #9
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    Re: roots

    thanks guys. God Bless
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