For polynomial of degrees two there is Quadratic Formula-Wiki Ref.. For degree three (as in your case), there is Closed-Form for Polynomial Degree 3-Cardano . This formula is not usually used maybe because it is difficult to apply and may result not-so-precise results (when used by machine) due to the square roots. but it gives the answer ...

2015年8月12日 · Given a cubic polynomial with real coefficients of the form $f(x) = Ax^3 + Bx^2 + Cx + D$ $(A \\neq 0)$ I am trying to determine what the necessary conditions of the ...

2018年4月2日 · For if this cubic polynomial has three real roots, but it is irreducible over $\mathbb{Q}$, then Cardano's Formula is going to force us to consider cube roots of complex numbers. Which scarcely seems applicable to peoples that really only considered positive real …

While they do start getting awkward quickly, the next few ordinals are fairly well-defined, largely because of their occasional usage in solving cubic and quartic equations and in defining algebraic curves and surfaces: the Sextic, the Septic, and the Octic. Beyond that, they just don't show up often enough to be worth explicitly naming.

You can solve cubics using a similar idea to 'completing the square'. The first step is to note that $(x+y)^3=x^3+3x^2y+3xy^2+y^3$ and use this to remove the quadratic factor.

2014年11月5日 · The two most important reasons to use cubic splines instead of quadratic: Cubic splines are C2-continuous, which is handy sometimes; Quadratic splines "ring" a lot. If you move one of the data points, it will make big changes to the whole curve all the way out to the ends.

Prove irreducible cubic polynomial over $\mathbb{Q}$ with a cyclic galois group has real roots Hot Network Questions Why do mDNS packets reach my device with a subnet mask of 255.255.255.255?

Find a cubic polynomial whose graph has horizontal tangents at (−2, 5) and (2, 3)

2015年7月16日 · The Routh-Hurwitz Criterion gives a condition for roots lying in the open left half-plane for an arbitrary polynomial with complex coefficients which helps a little, but this criterion doesn't help me when the complex roots lie in the right half plane.

For the first, as Avatar described, you use that $\frac{1}{\alpha+1}$ satisfies the polynomial relation $6y^3-7y^2+3y-1=0$ to express its third power in lower powers of itself.