Show that $x - 1$ is a factor of $3x^{3} + 8x^{2} - 5x - 6$. Hence solve the equation $3x^{3} + 8x^{2} - 5x = 6$ completely. Hence solve the equation $3(y - 2)^{3} + 8(y - 2)^{2} - 5y + 4 = 0$.

1. Define the polynomial and substitute $x = 1$ to apply the Factor Theorem

Let $P(x) = 3x^3 + 8x^2 - 5x - 6$.
Substitute $x = 1$ into $P(x)$:
$P(1) = 3(1)^3 + 8(1)^2 - 5(1) - 6$

2. Evaluate the expression and conclude

$P(1) = 3 + 8 - 5 - 6 = 0$
Since $P(1) = 0$, by the Factor Theorem, $x - 1$ is a factor of $3x^3 + 8x^2 - 5x - 6$.

1. Rearrange the equation into the form $f(x) = 0$ to use the result from part (a)

$3x^{3} + 8x^{2} - 5x - 6 = 0$

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

Since $(x - 1)$ is a factor from part (a):
$3x^{3} + 8x^{2} - 5x - 6 = (x - 1)(Ax^{2} + Bx + C)$
By comparing coefficients:
$A = 3$
$C = 6$
Comparing coefficients of $x^{2}$:
$B - A = 8 \implies B - 3 = 8 \implies B = 11$
So, the quadratic factor is $3x^{2} + 11x + 6$.

3. Factorise the quadratic expression

$3x^{2} + 11x + 6 = (3x + 2)(x + 3)$

4. Solve the completely factorised equation

$(x - 1)(3x + 2)(x + 3) = 0$
$x = 1$, $x = -\frac{2}{3}$, or $x = -3$

1. Rewrite the equation to match the polynomial from part (b)

The given equation is $3(y - 2)^{3} + 8(y - 2)^{2} - 5y + 4 = 0$.
Notice that $-5y + 4 = -5(y - 2) - 10 + 4 = -5(y - 2) - 6$.
So the equation can be written as:
$3(y - 2)^{3} + 8(y - 2)^{2} - 5(y - 2) - 6 = 0$

2. Use the roots from part (b) to solve for $y$

Let $x = y - 2$. The equation becomes $3x^{3} + 8x^{2} - 5x - 6 = 0$.
From part (b), the roots are $x = 1$, $x = -\frac{2}{3}$, and $x = -3$.
Therefore, $y - 2 = 1 \implies y = 3$
$y - 2 = -\frac{2}{3} \implies y = \frac{4}{3}$
$y - 2 = -3 \implies y = -1$