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

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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
hum i now this!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
this can not be factored!

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Other answers:

@RadEn indeed this is not reducible over the reals... check the discriminant.
actually the discriminant used in quadratic equation, not for cubic eq
Correct @RadEn
@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... 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. 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.
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.
@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.
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!!!!!!
@lordcyborg no argument here! Just discussing math :-p

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