如图所示为 $y = x^{3} - 3x^{2} - x - 12$ 的图像。

1. Evaluate the polynomial at $x = 4$ using the Factor Theorem

Let $P(x) = x^3 - 3x^2 - x - 12$.
$P(4) = (4)^3 - 3(4)^2 - (4) - 12$
$P(4) = 64 - 48 - 4 - 12$
$P(4) = 0$

1. Find the quadratic factor of the cubic expression using the linear factor from part (ii)

By long division or comparing coefficients:
$x^3 - 3x^2 - x - 12 = (x - 4)(x^2 + bx + c)$
Comparing $x^2$ coefficients: $-4 + b = -3 \implies b = 1$
Comparing constants: $-4c = -12 \implies c = 3$
So, $x^3 - 3x^2 - x - 12 = (x - 4)(x^2 + x + 3)$

2. Substitute the factorised form into the equation and rearrange

$(x - 4)(x^2 + x + 3) = 5(x - 4)$
$(x - 4)(x^2 + x + 3) - 5(x - 4) = 0$
$(x - 4)(x^2 + x + 3 - 5) = 0$
$(x - 4)(x^2 + x - 2) = 0$

3. Factorise the resulting quadratic and solve for $x$

$(x - 4)(x + 2)(x - 1) = 0$
$x = 4$, $x = -2$, or $x = 1$