1. Find the roots of the quadratic

Given $x^2+2x+2=0$

Using the quadratic formula, $x=\frac{-2\pm\sqrt{4-8}}{2}=\frac{-2\pm\sqrt{-4}}{2}=-1\pm i$

So the roots are $\alpha=-1+i,\qquad \beta=-1-i$

2. Write the roots in polar form

For $\alpha=-1+i$:

• Modulus: $|\alpha|=\sqrt{(-1)^2+1^2}=\sqrt{2}$
• Argument: point $(-1,1)$ lies in quadrant II, so $\arg(\alpha)=\frac{3\pi}{4}$

Hence, $\alpha=\sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$

Similarly, $\beta=\sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right)+i\sin\left(-\frac{3\pi}{4}\right)\right)$

3. Compute $\alpha^{15}$ and $\beta^{15}$ using De Moivre’s theorem

$\alpha^{15}=(\sqrt{2})^{15}\left(\cos\frac{45\pi}{4}+i\sin\frac{45\pi}{4}\right)$ $\beta^{15}=(\sqrt{2})^{15}\left(\cos\left(-\frac{45\pi}{4}\right)+i\sin\left(-\frac{45\pi}{4}\right)\right)$

Now, $(\sqrt{2})^{15}=2^{15/2}=2^7\sqrt{2}=128\sqrt{2}$

Also, $\frac{45\pi}{4}=11\pi+\frac{\pi}{4}$ So, $\cos\frac{45\pi}{4}=\cos\left(11\pi+\frac{\pi}{4}\right)=-\cos\frac{\pi}{4}=-\frac{\sqrt{2}}{2}$ $\sin\frac{45\pi}{4}=\sin\left(11\pi+\frac{\pi}{4}\right)=-\sin\frac{\pi}{4}=-\frac{\sqrt{2}}{2}$

Thus, $\alpha^{15}=128\sqrt{2}\left(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)=128(-1-i)$

Similarly, $\beta^{15}=128(-1+i)$

4. Add them

$\alpha^{15}+\beta^{15}=128(-1-i)+128(-1+i)$ $=128(-2)=-256$

5. Match with the options

The correct value is $\boxed{-256}$ So the correct option is A.