(a)

1. Apply the Remainder Theorem to find the remainders for both polynomials when divided by $x+2$.

By the Remainder Theorem, the remainders are $P(-2)$ and $Q(-2)$.
$P(-2) = a(-2)^3 + (-2)^2 + b(-2) + 10$
$P(-2) = -8a + 4 - 2b + 10 = -8a - 2b + 14$

$Q(-2) = (-2)^3 + a(-2)^2 - 5(-2) + b$
$Q(-2) = -8 + 4a + 10 + b = 4a + b + 2$

2. Equate the remainders and simplify to show the required equation.

Since the remainders are the same:
$-8a - 2b + 14 = 4a + b + 2$
$12a + 3b = 12$
Dividing by 3 throughout:
$4a + b = 4$ (shown)

1. Find the derivative $P'(x)$

$P(x) = ax^3 + x^2 + bx + 10$
$P'(x) = 3ax^2 + 2x + b$

2. Apply the Factor Theorem to $P'(x)$

Since $3x - 2$ is a factor of $P'(x)$, $P'\left(\frac{2}{3}\right) = 0$.
$3a\left(\frac{2}{3}\right)^2 + 2\left(\frac{2}{3}\right) + b = 0$
$3a\left(\frac{4}{9}\right) + \frac{4}{3} + b = 0$
$\frac{4}{3}a + \frac{4}{3} + b = 0$
$4a + 4 + 3b = 0$
$4a + 3b = -4$

3. Solve the simultaneous equations for $a$ and $b$

From part (a)(i), we have $4a + b = 4$.
Subtracting this from $4a + 3b = -4$:
$(4a + 3b) - (4a + b) = -4 - 4$
$2b = -8 \implies b = -4$
Substitute $b = -4$ into $4a + b = 4$:
$4a - 4 = 4 \implies 4a = 8 \implies a = 2$

1. Find a linear factor of the polynomial using the Factor Theorem.

Let $f(x) = 2x^3+4x^2-5x-7$.
$f(-1) = 2(-1)^3 + 4(-1)^2 - 5(-1) - 7 = -2 + 4 + 5 - 7 = 0$
By the Factor Theorem, $(x+1)$ is a factor.

2. Factorise the cubic polynomial into a linear and a quadratic factor.

$2x^3+4x^2-5x-7 = (x+1)(2x^2 + cx - 7)$
Comparing coefficients of $x^2$:
$c + 2 = 4 \implies c = 2$
So, $2x^3+4x^2-5x-7 = (x+1)(2x^2 + 2x - 7)$

3. Solve the quadratic equation $2x^2 + 2x - 7 = 0$ using the quadratic formula.

$2x^2 + 2x - 7 = 0$
$x = \frac{-2 \pm \sqrt{2^2 - 4(2)(-7)}}{2(2)}$
$x = \frac{-2 \pm \sqrt{4 + 56}}{4}$
$x = \frac{-2 \pm \sqrt{60}}{4}$

4. Simplify the surd to express the non-exact solutions in the required form.

$x = \frac{-2 \pm \sqrt{4 \times 15}}{4}$
$x = \frac{-2 \pm 2\sqrt{15}}{4}$
$x = -\frac{1}{2} \pm \frac{1}{2}\sqrt{15}$