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anonymous
 3 years ago
how do i find the zeros of f(x)=x^33x^23?
anonymous
 3 years ago
how do i find the zeros of f(x)=x^33x^23?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0find a zero by hit and trial method. Next u get a factor so divide by it and then find the rest of the zeroes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hum i now this!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.0this can not be factored! http://www.wolframalpha.com/input/?i=x%5E3++3x%5E2++3+%3D+0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@RadEn indeed this is not reducible over the reals... check the discriminant.

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.0actually the discriminant used in quadratic equation, not for cubic eq

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@RadEn the discriminant you are likely familiar with is \(\Delta=b^24ac\), 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 \(b^2c^24ac^34b^3d27a^2d^2+18abcd\), a homogeneous polynomial of degree 4.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Cubic_polynomial#The_nature_of_the_roots So let us check our discriminant now :)$$\Delta=b^2c^24ac^34b^3d27a^2d^2+18abcd$$Here, we have \(x^33x^23=0\) so \(a=1, b=3,c=0,d=3\). Plug them in:$$\begin{align*}\Delta&=27(1)^2(3)^24(1)(0)4(3)^3(3)27(1)^2(3)^2+18(1)(3)(0)(3)\\&=27\times904\times8127\times9+0\\&=4\times81\lt0\end{align*}$$Because \(\Delta\lt0\), we have only one real root and two complex conjugate roots.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmmmm @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!!!!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@lordcyborg no argument here! Just discussing math :p
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