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Math Help - How to factor x^5-5x^3+4x ?

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    How to factor x^5-5x^3+4x ?

    x^5-5x^3+4x

    ...I don't know exactly how to do that :-|
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  2. #2
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    Quote Originally Posted by echorobotics View Post
    x^5-5x^3+4x

    ...I don't know exactly how to do that :-|
    x^5 - 5x^3 + 4x = x(x^4 - 5x^2 + 4).

    Now you have a quadratic in x^2. So let X = x^2.

     = x(X^2 - 5X + 4)

     = x(X - 1)(X - 4)

     = x(x^2 - 1)(x^2 - 4)

    Now use difference of two squares

     = x(x^2 - 1^2)(x^2 - 2^2)

     = x(x - 1)(x + 1)(x - 2)(x + 2).
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  3. #3
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    Or.
    Take the possible zero's using the rational rooth theorem.
    \pm1,\pm2,\pm3,\pm4<br />
    1.Then used synthetic division a few times.
    2.Once done, you will have found all the zero's (where r=0).
    3.Then just put it in factored form.


    1.
    1) 1,0,-5,0,4
    1(1)+0=1(1)-5=-4*1+0=-4+4=0
    r=0
    -1) 1,1,-4,-4
    1(-1)+1=0(1)+(-4)=-4+4=0
    r=0

    2) 1,0,-4,0
    1(2)+0=2(2)+(-4)=0(2)+0=0
    r=0

    -2) 1,2,0,0
    1(-2)+2=0
    r=0

    3.Now look at the coefficients left.

    1,0,0,0
    subtract leading exponent by number of divisions made::

    5-4=1
    therefore the coefficient is x.

    place the zero's in factored form
    (x+1)(x-1)(x+2)(x-2)

    multiply by x

    x(x+1)(x-1)(x+2)(x-2)


    Seems harder & longer, but sometimes this method can be a life saver >.> for me at least

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