Given that $\text{f}(x) = x^3 + 3x^2 - 4$, show that $x+2$ is a factor of $\text{f}(x)$, and hence factorise $\text{f}(x)$ completely. Hence, express $\frac{2x^3 + 6x^2 + 7}{x^3 + 3x^2 - 4}$ in partial fractions.

1. Use the Factor Theorem to show that $x+2$ is a factor of $\text{f}(x)$.

$\text{f}(-2) = (-2)^3 + 3(-2)^2 - 4$
$= -8 + 3(4) - 4$
$= -8 + 12 - 4 = 0$
Since $\text{f}(-2) = 0$, by the Factor Theorem, $x+2$ is a factor of $\text{f}(x)$.

2. Find the quadratic factor by dividing $\text{f}(x)$ by $x+2$.

Let $\text{f}(x) = (x+2)(Ax^2 + Bx + C)$.
Comparing coefficients of $x^3$: $A = 1$.
Comparing constant terms: $2C = -4 \implies C = -2$.
Comparing coefficients of $x^2$: $2A + B = 3 \implies 2(1) + B = 3 \implies B = 1$.
So, $\text{f}(x) = (x+2)(x^2 + x - 2)$.

3. Factorise the quadratic factor completely.

$x^2 + x - 2 = (x+2)(x-1)$
Therefore, $\text{f}(x) = (x+2)(x+2)(x-1) = (x+2)^2(x-1)$.

1. Recognise that the fraction is improper and perform long division

The degree of the numerator (3) is equal to the degree of the denominator (3), so the fraction is improper.
$\frac{2x^3 + 6x^2 + 7}{x^3 + 3x^2 - 4} = \frac{2(x^3 + 3x^2 - 4) + 8 + 7}{x^3 + 3x^2 - 4} = 2 + \frac{15}{x^3 + 3x^2 - 4}$

2. Use the factorised form of the denominator from part (i)

From part (i), $x^3 + 3x^2 - 4 = (x-1)(x+2)^2$.
So, we need to express $\frac{15}{(x-1)(x+2)^2}$ in partial fractions.

3. Set up the partial fraction decomposition

$\frac{15}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}$

4. Form the basic equation to solve for constants

$15 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)$

5. Solve for A, B, and C

Let $x = 1$:
$15 = A(3)^2 \implies 9A = 15 \implies A = \frac{5}{3}$

Let $x = -2$:
$15 = C(-3) \implies C = -5$

Compare coefficients of $x^2$:
$0 = A + B \implies B = -A = -\frac{5}{3}$

6. Write the final expression

$\frac{2x^3 + 6x^2 + 7}{x^3 + 3x^2 - 4} = 2 + \frac{5}{3(x-1)} - \frac{5}{3(x+2)} - \frac{5}{(x+2)^2}$