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