From the given information, how do we use the Remainder Theorem to reach the answer? Why does the sequence for the polynomial alternate when it equals $-30$, and not alternate when it equals $10$?  Why is the denominator when the sum of the geometric series is equal to $30$, $1+r$, and not $1-r$?  I am unable to correctly relate the Remainder Theorem to its use in the given answer.
The question:
When a polynomial of degree $3$ is written in descending powers of $x$ the coefficients of its four terms form a geometric sequence.
When the polynomial is divided by $x+1$ it s remainder is $-30$.  Then it is divided by $x-1$ its remainder is $10$.
The answer:
Use the Remainder Theorem.  We have
$-a+ar-ar^2+ar^3=-30$
and
$a+ar+ar^2+ar^3=10$
Solving simultaneously:
$\frac{a(1-r^4)}{1+r}=30$
and
$\frac{a(1-r^4)}{1-r}=10$
Giving $r=-\frac{1}{2}$ and $a=16$
The polynomial is therefore $16x^3-8x^2+4x-2$
 A: The polynomial is a cubic whose coefficients are a geometric sequence (in the order of descending powers, i.e. starting with $x^3$ and ending with the constant term), meaning that our polynomial must take the form
$$p(x) = ax^3 + arx^2 + ar^2x + ar^3.$$
Remainder theorem states that the remainder of $p(x)$ when divided by $x - \alpha$, is $p(\alpha)$.
So, when dividing $p(x)$ by $x - 1$ (i.e. $\alpha = 1$), the remainder is $p(1)$. Thus,
$$10 = p(1) = a1^3 + ar1^2 + ar^21 + ar^3 = a + ar + ar^2 + ar^3.$$
Similarly, when dividing $p(x)$ by $x + 1$ (i.e. $\alpha = -1$), the remainder is $p(-1)$. Thus,
$$-30 = p(-1) = a(-1)^3 + ar(-1)^2 + ar^2(-1) + ar^3 = -a + ar - ar^2 + ar^3.$$
This should hopefully explain the alternating signs. The next observation you need to make is that both of the above expressions are geometric series. Each have $4$ terms. The first has an initial term of $a$ and a common ratio of $r$, while the second has an initial term of $-a$ and a common ratio of $-r$. The first one is precisely the usual formula for geoemetric series:
$$10 = a + ar + ar^2 + ar^3 = a\frac{1 - r^4}{1 - r}.$$
To get a formula for the second one, we simply replace $r$ with $-r$ and $a$ with $-a$, to get
$$-30 = -a + ar - ar^2 + ar^3 = (-a)\frac{1 - (-r)^4}{1 - (-r)} = -a\frac{1 - r^4}{1 + r}.$$
To get the equation from the solution, simply multiply both sides by $-1$.
