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Math Help - Factorize

  1. #1
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    Factorize

    How do i factorize this?

    <br />
r^4-4r^3+7r^2-6r+2=0<br />
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  2. #2
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    Quote Originally Posted by MilK View Post
    How do i factorize this?

    <br />
r^4-4r^3+7r^2-6r+2=0<br />
    It is (r-1)^2(r^2-2r+2)=0.
    You can see that r=1 is a solution so you get (r-1)(r^3-3r^2+4r-2)=0 and you can realize that r=1 is a root of r^3-3r^2+4r-2=0 too, so you get (r-1)^2(r^2-2r+2)=0. And r^2-2r+2=0 hasn't got any real roots (because D=2^2-4*2<0) so you can't factorize that.
    Last edited by james_bond; March 8th 2008 at 09:24 AM.
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  3. #3
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    Quote Originally Posted by james_bond View Post
    It is (r-1)^2(r^2-2r+2)=0.
    You can see that r=1 is a solution so you get (r-1)(r^3-3r^2+4r-2)=0 and you can realize that r=1 is a root of r^3-3r^2+4r-2=0 too, so you get (r-1)^2(r^2-2r+2). And r^2-2r+2=0 hasn't got any real roots (because D=2^2-4*2<0) so you can't factorize that.
    Sorry i am not gd at realization, is there any other way?
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  4. #4
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    Hello, MilK!

    How do i factorize this?

    r^4-4r^3+7r^2-6r+2\:=\:0
    I factored this at your other post,
    . . but here's how I did it.


    Knowing the Rational Roots Theorem,
    . . I knew that the only rational roots are: . \pm1,\:\pm2

    I tried r = 1:\;\;1^4 - 4(1^3) + 7(1^2) - 6(1) + 2 \;=\;0

    From the Remainder Theorem, I knew that (r-1) is a factor.


    Using long division: . (r - 1)(r^3-3r^2+4r-2)

    I saw that r = 1 is a root of the cubic.
    . . Hence, (r-1) is a factor.

    And we have: . (r-1)(r-1)(r^2 -2r + 2)


    The quadratic does not factor; use the Quadratic Formula.

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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MilK View Post
    Sorry i am not gd at realization, is there any other way?
    Have you heard of the rational root theorem?

    -Dan
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  6. #6
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    okay i got it, thanks everyone
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