Respected All
I got stuck in it and need your help.
We know that if $\alpha_1, \cdots, \alpha_5$ be the roots of $p(x):=x^5+ax^4+bx^3+cx^2+dx+e=0$ then the equation which roots are opposite in sign viz $-\alpha_1, \cdots, -\alpha_5$ can be determined by the substitution $y=-x$. And deduction says that we just have to change the signs in alternate coeffecients of $p(x)=0$.
I was thinking what will be the equation if the roots of it be $\alpha_1, \alpha_2, -\alpha_3,-\alpha_4,-\alpha_5$ ? Suppose that the required polynomial is $q(x):=x^5+Ax^4+Bx^3+Cx^2+Dx+E$. Can we get the values of $A, B, C, D, E$ in terms of $a, b, c, d, e$ ?
Direct substitution is not working here. What shall I do here ?
Please show me the path.
Thanking you in advance
P.s. Could not find the proper tag for it. Some one please edit if necessary
Roots of polynomials are an unordered set, so you can only do transformations of coefficients that are invariant to the permutation of roots. You can, for instance, negate all roots by appropriately changing the signs of coefficients, but you cannot even specify what you require if you want to only negate a subset of roots, unless you already have the roots. For instance, you can say you want to negate two roots with the largest magnitude (or something similar) but this requires you to actually find the roots. Because of this, your question is related to the Galois theory and the inability of extracting roots of general higher order polynomials. In fact, the insensitivity of coefficients to permutation of roots is at the core of the proof that polynomials with order 5 or greater cannot be solved in radicals, unless they have a special separable form.