Re: Polynomial sub problem

No, I don't think that works. We know that

$\displaystyle f(y) = f(x^2) = (x^2)^3 + a(x^2)^2 + bx^2 + c$

Letting $\displaystyle x = \sqrt{Q}$ and $\displaystyle y = Q$ yields a root that we already know...

Re: Polynomial sub problem

oh, i mean by sub. y=x^{2}, then x^{3}+ax^{2}+bx+c = 0 becomes

x(x^{2}+b) = -(a^{2}+c)

x(y+b)=-(ay+c)

x^{2}(y+b)^{2}=(ay+c)^{2}

y(y^{2}+2by+b^{2}) = (a^{2}y^{2}+2acy+c^{2})

y^{3}+2by^{2}+b^{2}y = a^{2}y^{2} + 2acy + c^{2}

y^{3}+y^{2}(2b-a^{2})+y(b^{2}-2ac)-c^{2} =0

then eqaution y^{3}+y^{2}(2b-a^{2})+y(b^{2}-2ac)-c^{2} =0, has required root, sorry for the misleading of using same function notication

Re: Polynomial sub problem

You're way overthinking this. Also, your first step:

$\displaystyle x(x^2 + b) = -(a^2 + c)$

It should be

$\displaystyle x(x^2 + b) = -(ax^2 + c)$

Re: Polynomial sub problem

oh ya, sorry for my careless mistake... but why it works actually ><?