The remainder when $x^3 + 17x^2 + 7x - 9$ is divided by $x - a$ is twice the remainder when it is divided by $x + a$.

1. Find the expressions for the remainders using the Remainder Theorem.

Let $P(x) = x^3 + 17x^2 + 7x - 9$.
When divided by $x - a$, the remainder is $P(a) = a^3 + 17a^2 + 7a - 9$.
When divided by $x + a$, the remainder is $P(-a) = (-a)^3 + 17(-a)^2 + 7(-a) - 9 = -a^3 + 17a^2 - 7a - 9$.

2. Set up the equation using the given condition.

Since the first remainder is twice the second remainder:
$P(a) = 2P(-a)$
$a^3 + 17a^2 + 7a - 9 = 2(-a^3 + 17a^2 - 7a - 9)$

3. Simplify the equation to show the required result.

$a^3 + 17a^2 + 7a - 9 = -2a^3 + 34a^2 - 14a - 18$
$a^3 + 2a^3 + 17a^2 - 34a^2 + 7a + 14a - 9 + 18 = 0$
$3a^3 - 17a^2 + 21a + 9 = 0$ (shown)

1. Use the Factor Theorem to find a linear factor of the cubic polynomial.

Let $P(a) = 3a^3 - 17a^2 + 21a + 9$.
Testing factors of 9:
$P(3) = 3(3)^3 - 17(3)^2 + 21(3) + 9$
$P(3) = 81 - 153 + 63 + 9 = 0$
By the Factor Theorem, $(a - 3)$ is a factor.

2. Find the quadratic factor by comparing coefficients or polynomial division.

$3a^3 - 17a^2 + 21a + 9 = (a - 3)(3a^2 + ba - 3)$
Comparing coefficients of $a^2$:
$-9 + b = -17 \implies b = -8$
So the quadratic factor is $3a^2 - 8a - 3$.

3. Factorise the quadratic factor.

$3a^2 - 8a - 3 = (3a + 1)(a - 3)$

4. Solve the equation to find all values of $a$.

$(a - 3)(3a + 1)(a - 3) = 0$
$(a - 3)^2(3a + 1) = 0$
$a = 3$ or $a = -\frac{1}{3}$