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lordcyborg Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

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

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

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

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

lordcyborg Group TitleBest ResponseYou've already chosen the best response.0
Correct @RadEn
 one year ago

oldrin.bataku Group TitleBest 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.
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.0
http://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.
 one year ago

lordcyborg Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

oldrin.bataku Group TitleBest 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.
 one year ago

lordcyborg Group TitleBest ResponseYou've already chosen the best response.0
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!!!!!!
 one year ago

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