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CJG123

  • 3 years ago

how do i find the zeros of f(x)=x^3-3x^2-3?

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  1. lordcyborg
    • 3 years ago
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    find a zero by hit and trial method. Next u get a factor so divide by it and then find the rest of the zeroes

  2. AKANicole
    • 3 years ago
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    hum i now this!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  3. RadEn
    • 3 years ago
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    this can not be factored! http://www.wolframalpha.com/input/?i=x%5E3+-+3x%5E2+-+3+%3D+0

  4. oldrin.bataku
    • 3 years ago
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    @RadEn indeed this is not reducible over the reals... check the discriminant.

  5. RadEn
    • 3 years ago
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    actually the discriminant used in quadratic equation, not for cubic eq

  6. lordcyborg
    • 3 years ago
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    Correct @RadEn

  7. oldrin.bataku
    • 3 years ago
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    @RadEn the discriminant you are likely familiar with is \(\Delta=b^2-4ac\), and it is (indeed) only for quadratic equations \(ax^2+bx+c=0\). But the discriminant as a concept is extended to polynomials of other degree... http://en.wikipedia.org/wiki/Discriminant For example, a cubic equation \(ax^3+bx^2+cx+d=0\) yields a discriminant \(b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd\), a homogeneous polynomial of degree 4.

  8. oldrin.bataku
    • 3 years ago
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    http://en.wikipedia.org/wiki/Cubic_polynomial#The_nature_of_the_roots So let us check our discriminant now :-)$$\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$Here, we have \(x^3-3x^2-3=0\) so \(a=1, b=-3,c=0,d=-3\). Plug them in:$$\begin{align*}\Delta&=27(1)^2(-3)^2-4(1)(0)-4(-3)^3(-3)-27(1)^2(-3)^2+18(1)(-3)(0)(-3)\\&=27\times9-0-4\times81-27\times9+0\\&=-4\times81\lt0\end{align*}$$Because \(\Delta\lt0\), we have only one real root and two complex conjugate roots.

  9. lordcyborg
    • 3 years ago
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    No need of such big discriminants coz one can always use hit and trial method to find the zero of a simple equation like this..... and then use the discriminant for the quadratic equation after creating a factor with that zero.

  10. oldrin.bataku
    • 3 years ago
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    @lordcyborg the real root is not trivial here, so tell me more about how you plan on using a simple "hit and trial" method :-) In fact, this polynomial's only real root isn't rational.

  11. lordcyborg
    • 3 years ago
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    hmmmm @oldrin.bataku I did not take a good look at d equation. So I suggested to use just hit and trial method. Yup u r correct. Sorry for arguing!!!!!!

  12. oldrin.bataku
    • 3 years ago
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    @lordcyborg no argument here! Just discussing math :-p

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