if $\alpha,\beta,\gamma,\delta$ are the roots of equation $x^4+4x^3-6x^2+7x-9$ then find $\prod(1+\alpha^2)$ I know
I know the sum and products and other things about the roots but I am not being to rearrange them such that I am able to get the required answer
On
Your polynomial is irreducible over $\mathbb{Q}$ since it is irreducible over $\mathbb{F}_2$, hence it is the minimal polynomial of $\alpha$. The minimal polynomial of $\alpha^2$ is given by
$$ \left[\underbrace{\left(x^4-6x^2-9\right)^2}_{\text{even part}}-\underbrace{\left(4x^3+7x\right)^2}_{\text{odd part}}\right]_{x\mapsto \sqrt{x}} = x^4-28x^3-38x^2+59x+81=g(x)$$ and $$ \prod_\text{cyc}(1+\alpha^2) = g(-1) =\color{red}{13}.$$ As an alternative, the LHS is the determinant of $M^2+I$ where $M$ is the companion matrix of the original polynomial.
On
Another way is to write it as $\;\prod(1+\alpha^2) = \prod\big((1+i\alpha) \cdot (1-i\alpha)\big)=\prod(1+i\alpha) \cdot \prod(1-i\alpha)\,$.
If $\,\alpha\,$ are the roots of $\,P(x)\,$, then the polynomial $\,Q(z)\,$ with roots $\,1+i \alpha\,$ is the result of the substitution $\,1+ix = z$ $\iff x = i(1-z)\,$. After routine calculations:
$$ Q(z) = P\big(i(1-z)\big) = z^4 - 4(1 - i) z^3 + 12(1 - i) z^2 - (16 - 5 i) z - (2 - 3 i) $$
It follows from Vieta's relations that $\,\prod (1+i \alpha) = -(2-3i)\,$.
By a similar argument $\,\prod (1-i \alpha) = -(2+3i)\,$, so in the end $\,\prod(1+\alpha^2)= |2-3i|^2=13\,$.
Transforming roots from $\alpha$ to $1+\alpha^2$,
$$y=1+\alpha^2$$
$$\alpha=\sqrt{y-1}$$
As $\alpha$ satisfies our given polynomial, Substitute $\alpha$ into that.
We get new equation with roots $1+\alpha^2,1+\beta^2,\ldots$
$$(x-1)^2+4(x-1)\sqrt{x-1}-6(x-1)+7\sqrt{x-1}-9=0$$
Rearranging and squaring,
$$(x-1)^2(x-7)^2+81-18(x-1)(x-7)=(x-1)(4x+3)^2$$
$$(x^2+1-2x)(x^2+49-14x)+81-18(x^2+7-8x)=(x-1)(16x^2+9+24x)$$
$$x^4-32x^3+34x^2+45x+13=0$$
$$\text{Product of roots}=\frac{13}{1}=13$$