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Math Help - Finding Roots of Polynomials

  1. #1
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    Finding Roots of Polynomials

    On problems like  -12x^3+12x^2+24x

    I try to solve for roots.
    What I get is, (12+x)(-x^2+x+2), leaving me to find

    x=-12 x=-1 x=2

    Come to find out, x=-12 should be x=0. How can I fix this?

    Also, for problems that do not factor so easy,  -32x^3+12x+0 , what are some factor methods? thank you
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  2. #2
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    Quote Originally Posted by skyslimit View Post
    On problems like  -12x^3+12x^2+24x

    I try to solve for roots.
    What I get is, (12+x)(-x^2+x+2), leaving me to find

    x=-12 x=-1 x=2

    Come to find out, x=-12 should be x=0. How can I fix this?

    Also, for problems that do not factor so easy,  -32x^3+12x+0 , what are some factor methods? thank you
    It looks like it should be what you said except for one minor mistake, but heres a tip that will help you with the first and 2nd making life easier.

    Ok first off:

     -12x^3+12x^2+24x

    Recognize the common factor here 12x, personally I would pull out a -12x

    thus becoming (-12x)(x^2-x-2) this become an easily visible factor(depending on your math level) : we find ourselves with 3 equations :

    12x x+1 x-2

    I don't see where you got the answer -12, I see 0 though,
    Pulling out an 12+x gives you a completely different equation then the original, wouldn't it.. Im no professional but I can factor and I don't see how -12 could be an answer
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  3. #3
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    <br /> <br />
-32x^3 + 12x<br />

    Take out the common factor of -6x

    <br /> <br />
-6x(4x^2 - 2)<br />

    Then we get -6x = 0
    Therefore x=0

    Move 2 over to the other side

    <br /> <br />
4x^2 = 2<br />

    Divide both sides by 4

    <br /> <br />
x^2 = \frac{1}{2}<br />

    Then take the square root of both sides and you get 2 more roots

    <br /> <br />
x = + \sqrt \frac{1}{2}<br /> <br />

    <br /> <br />
x = - \sqrt \frac{1}{2}<br />
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  4. #4
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    Further Examination

    If you plug in -12 to the original equation it equals 22,176.. which is definately not a root haha, did your professor/teacher tell you that was the answer or is it on answer? Many of my personal worksheets for Calculus have wrong answers on them all the time, teachers update questions but forget to update the answers.

    Thanks for the thanks
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by iz1hp View Post
    <br /> <br />
-32x^3 + 12x<br />

    Take out the common factor of -6x

    <br /> <br />
-6x(4x^2 - 2)<br />

    Then we get -6x = 0
    Therefore x=0

    Move 2 over to the other side

    <br /> <br />
4x^2 = 2<br />

    Divide both sides by 4

    <br /> <br />
x^2 = \frac{1}{2}<br />

    Then take the square root of both sides and you get 2 more roots

    <br /> <br />
x = + \sqrt \frac{1}{2}<br /> <br />

    <br /> <br />
x = - \sqrt \frac{1}{2}<br />
    You can't factor out a 6

    Here is the solution
    http://www.mathhelpforum.com/math-he...s-extrema.html
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