1. Factor the polynomial

Given $x^7+3x^5-13x^3-15x=0$

First take out $x$: $x(x^6+3x^4-13x^2-15)=0$

Now let $y=x^2$. Then we need to factor $y^3+3y^2-13y-15$

Try integer roots. For $y=3$: $3^3+3(3^2)-13(3)-15=27+27-39-15=0$ So $(y-3)$ is a factor.

Divide: $y^3+3y^2-13y-15=(y-3)(y^2+6y+5)$ And $y^2+6y+5=(y+1)(y+5)$

Hence $x^6+3x^4-13x^2-15=(x^2-3)(x^2+1)(x^2+5)$

Therefore $x(x^2-3)(x^2+1)(x^2+5)=0$

So the 7 roots are: $0,\\ \,\pm \sqrt3,\ \pm i,\ \pm i\sqrt5$

2. Arrange roots by decreasing modulus

Their moduli are:

• $|\pm i\sqrt5|=\sqrt5$
• $|\pm \sqrt3|=\sqrt3$
• $|\pm i|=1$
• $|0|=0$

Thus, in the order $|\alpha_1|\ge |\alpha_2|\ge \cdots \ge |\alpha_7|$ we must have:

• $\alpha_1,\alpha_2=\pm i\sqrt5$
• $\alpha_3,\alpha_4=\pm \sqrt3$
• $\alpha_5,\alpha_6=\pm i$
• $\alpha_7=0$

Within each equal-modulus pair, order does not matter, but the product of each pair is fixed.

3. Compute the required expression

We need $\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6$

Now, $\alpha_1\alpha_2=(i\sqrt5)(-i\sqrt5)=5$ $\alpha_3\alpha_4=(\sqrt3)(-\sqrt3)=-3$ $\alpha_5\alpha_6=(i)(-i)=1$

Therefore,