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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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! http://www.wolframalpha.com/input/?i=x%5E3+-+3x%5E2+-+3+%3D+0

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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... 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.
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.
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question