Let's go through each part of the problem step by step.

Part (a)

Given that $(x - 1)$ is a factor of $f(x)$, show that $a + b + c = -1$.

Since $(x - 1)$ is a factor of $f(x)$, then $f(1) = 0$. Substituting $x = 1$ into the function:

$f(1) = 1^3 + a \cdot 1^2 + b \cdot 1 + c = 1 + a + b + c = 0.$

Therefore,

$a + b + c = -1.$

Part (b)

Find the values of $a$, $b$, and $c$.

From the problem, we are given the following conditions:

1. When $f(x)$ is divided by $(x - 2)$, the remainder is $-4$. This implies $f(2) = -4$.

   Substituting $x = 2$:

   $f(2) = 2^3 + a \cdot 2^2 + b \cdot 2 + c = 8 + 4a + 2b + c = -4.$

   Simplifying, we get:

   $4a + 2b + c = -12. \quad \text{(Equation 1)}$

2. When $f(x)$ is divided by $(x - 3)$, the remainder is $-12$. This implies $f(3) = -12$.

   Substituting $x = 3$:

   $f(3) = 3^3 + a \cdot 3^2 + b \cdot 3 + c = 27 + 9a + 3b + c = -12.$

   Simplifying, we get:

   $9a + 3b + c = -39. \quad \text{(Equation 2)}$

Now, we have three equations:

1. $a + b + c = -1$ (from part a).
2. $4a + 2b + c = -12$ (Equation 1).
3. $9a + 3b + c = -39$ (Equation 2).

Let's solve these equations:

Subtract equation 1 from equation 2:

$(9a + 3b + c) - (4a + 2b + c) = -39 - (-12),$

which simplifies to:

$5a + b = -27. \quad \text{(Equation 3)}$

Subtract equation 1 from equation 3:

$(4a + 2b + c) - (a + b + c) = -12 - (-1),$

which simplifies to:

$3a + b = -11. \quad \text{(Equation 4)}$

Now, subtract equation 4 from equation 3:

$(5a + b) - (3a + b) = -27 - (-11),$

which simplifies to:

$2a = -16,$

so:

$a = -8.$

Substitute $a = -8$ into equation 4:

$3(-8) + b = -11,$

which simplifies to:

$-24 + b = -11,$

so:

$b = 13.$

Substitute $a = -8$ and $b = 13$ into the original equation $a + b + c = -1$:

$-8 + 13 + c = -1,$

which simplifies to:

$5 + c = -1,$

so:

$c = -6.$

Thus, $a = -8$, $b = 13$, and $c = -6$.

Part (c)

Express $f(x)$ as the product of three linear factors.

We know $(x - 1)$ is a factor. Using the values $a = -8$, $b = 13$, and $c = -6$, the cubic function is:

$f(x) = x^3 - 8x^2 + 13x - 6.$

To factor $f(x)$, we divide by $(x - 1)$:

$f(x) = (x - 1)(x^2 - 7x + 6).$

Factoring the quadratic:

$x^2 - 7x + 6 = (x - 6)(x - 1).$

Thus:

$f(x) = (x - 1)^2 (x - 6).$

Part (d)

Sketch the graph of $f(x)$.

To sketch $f(x)$:

• Roots: $x = 1$ (with multiplicity 2) and $x = 6$.

• Y-intercept: Substitute $x = 0$:

  $f(0) = -6.$

Plot these points and sketch the curve with the shape expected from a cubic with a double root at $x = 1$ (tangent to the x-axis) and crossing at $x = 6$.

Would you like more details on any part? Here are some further explorations:

1. What happens if $(x - 2)$ was also a factor of $f(x)$?
2. How would the graph change if the coefficients were different?
3. Can you verify the factors of $f(x)$ using synthetic division?
4. How can the nature of roots be verified algebraically?
5. What if the remainder theorem is used on a quadratic function?

Tip: Always check for simple factors first, such as $x - 1$, when given roots or factors.