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lordcyborg
 one year 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

AKANicole
 one year ago
Best ResponseYou've already chosen the best response.0hum i now this!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

RadEn
 one year 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

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0@RadEn indeed this is not reducible over the reals... check the discriminant.

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

oldrin.bataku
 one year 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.

oldrin.bataku
 one year 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.

lordcyborg
 one year 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.

oldrin.bataku
 one year 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.

lordcyborg
 one year 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!!!!!!

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0@lordcyborg no argument here! Just discussing math :p
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