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Math Help - finding the roots

  1. #1
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    finding the roots

    what is the easiest way of finding the roots of these kind of equations (except guessing):

    x^4+x^3-7x^2-x+6=0

    or

    x^5-2x^4+2x^3-2x^2+x=0

    you can tell rigorously or sth easy. it's OK.
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  2. #2
    MHF Contributor red_dog's Avatar
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    a) x^4+x^3-7x^2-x+6=0

    Find integer roots. x_1=1, \ x_2=-1

    b) x(x^4-2x^3+2x^2-2x+1)=0

    x_1=0 and x^4-2x^3+2x^2-2x+1=0 is a reciprocal equation.
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  3. #3
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    Start by identifying that:

    x^4 + x^3 -7x^2 -x +6 = 0
    So x(x^3 + x^2 - 7x -1) = -6
    --> x^2(x^2 + x -7) = -5
    --> x^3(x +1) = 2
    --> x^4(1) = 1

    Therefore, your first x is going to be 1, and your second x is going to be -1.

    Next, use these roots to synthetically divide the equation down into something simpler or merely factor out the roots (x+1) and (x-1).

    You are left with (x+1)(x-1)(x^2 + x -6) which simplifies to (x+1)(x-1)(x+3)(x-2).
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  4. #4
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    Talking

    Quote Originally Posted by eyke View Post
    what is the easiest way of finding the roots of these kind of equations (except guessing):

    x^4+x^3-7x^2-x+6=0

    x^5-2x^4+2x^3-2x^2+x=0
    After looking at the graph in your graphing calculator, you can start by applying the Rational Roots Test to get a list of possible rational (fractional or whole-number) zeroes.

    Then you can try them (picking the more-likely options from the list by comparing with the calculator's picture) in synthetic division. Whichever values give you a "zero" remainder are valid roots.

    Then you apply the same process to whatever is remaining. The steps are illustrated here. :wink:
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