If I am given a polynomial of nth degree and asked to fond the product of real and imaginary roots what steps should I take?
I know how to calculate the sum or product of all roots of a polynomial of nth degree but how to separately find the product of real and imaginary roots?
For example if I'm asked to find the product of real roots of $2x^4 + 3x^3 - 11x^2 - 9x + 15 = 0$. What steps should I take?
This is too long for a comment. Just found a partial answer for $x^4+a_3x^3+a_2x^2+a_1x+a_0=0$. Let $y$ and $x$ denote the product of the real and complex roots and $b$ and $a$ the sums of the real and complex roots resp. Playing Vieta with the coefficients and those sums and product, you will find the following system of equations.
$$\begin{align*} yx&=a_0\\ a+b&=a_3\\ y+x+ab&=a_2\\ xb+ya&=a_1 \end{align*} $$
As this system is symmetric in $a\leftrightarrow b$ and in $x\leftrightarrow y$ we expect six solutions. For the given polynomial Mathematica confronts us with:
$$ \left\{\left\{a\to 0,b\to \frac{3}{2},x\to -3,y\to -\frac{5}{2}\right\},\left\{a\to \frac{3}{2},b\to 0,x\to -\frac{5}{2},y\to -3\right\},\left\{a\to \frac{1}{2} \left(5-2 \sqrt{3}\right),b\to \sqrt{3}-1,x\to -\frac{5 \sqrt{3}}{2},y\to -\sqrt{3}\right\},\left\{a\to -1-\sqrt{3},b\to \frac{1}{2} \left(5+2 \sqrt{3}\right),x\to \sqrt{3},y\to \frac{5 \sqrt{3}}{2}\right\},\left\{a\to \sqrt{3}-1,b\to \frac{1}{2} \left(5-2 \sqrt{3}\right),x\to -\sqrt{3},y\to -\frac{5 \sqrt{3}}{2}\right\},\left\{a\to \frac{1}{2} \left(5+2 \sqrt{3}\right),b\to -1-\sqrt{3},x\to \frac{5 \sqrt{3}}{2},y\to \sqrt{3}\right\}\right\} $$
Only the first solution fits to the given equation. At the moment I haven't a clue to de-symmetrize the conditions. Alas, it's a beginning.
Michael
To find the product of the roots of a polynomial use Vieta's formula which says if $\lbrace r_n \rbrace$ is the set of roots of an $n^{th}$ order polynomial $a_nx^n + a_{n-1}x^{n-1} + \dots +a_1x + a_0$ , then the product of the roots $r_1r_2 \dots r_n = (-1)^n\frac{a_0}{a_n}$. To find the sum of the roots you use the formula $\sum_{i=0}^{n}r_i = -\frac{a_{n-1}}{a_n} $
