EquationEnthusiast

It is impossible to eliminate both the $x^5$ term and the $x^6$ term at once, so we set $b = 0 \rightarrow f(x) + b \cdot g(x) = f(x) = 2x^5 - 6x^4 - 4x^3 + 12x^2 + 7x + 5$. $\text{deg}(f(x) + b \cdot g(x) = \boxed{5})$

Honestly, you're lucky to get complete solutions at all. We're doing your work for free, and more demands aren't exactly sufficient incentives for proffering full solutions.

Because $(i, i^2, i^3, i^4) = (i, -1, -i, 1)$, we have  $i + i^2 + i^3 \cdots + i^{258} + i^{259} = 64(i - 1 - i + 1) + i^{257} + i^{258} + i^{259}$. So, the desired sum is $i^{257} + i^{258} + i^{259} = i - 1 - i = \boxed{-1}$.

If $z = a + bi$, then $\overline{z} = a - bi$. So, $z \overline{z} = (a + bi)(a-bi) = a^2 + b^2$. Since $(a, b) = (3, 5)$, we have $z \overline{z} = 3^{2} + 5^{2} = 41.$

You published the same question twice. Both times, you failed to provide a diagram. As such, it is impossible for any of us to help you with your problem.

Hint: the number of ways to distribute $n$ indistinguishable balls into $k$ distinguishable boxes is found with the binomial coefficient $\binom{n + k - 1}{n}$.