The problem asks to model the distance of a ship from its harbour using the function $d(t) = -3t^3 + 3t^2 + 18t$, where $t$ is the time elapsed in hours since departure from the harbour.

Part a: Factor the time function.

We are tasked with factoring the function $d(t) = -3t^3 + 3t^2 + 18t$.

1. Factor out the greatest common factor (GCF):

The GCF of the terms $-3t^3$, $3t^2$, and $18t$ is $3t$. So, we factor out $3t$ from the expression:

$d(t) = 3t(-t^2 + t + 6)$

2. Factor the quadratic expression:

Now, we need to factor the quadratic expression inside the parentheses, $-t^2 + t + 6$.

To simplify factoring, let's rewrite the expression by factoring out $-1$:

$-t^2 + t + 6 = -(t^2 - t - 6)$

Now, factor $t^2 - t - 6$. We need two numbers that multiply to $-6$ and add to $-1$. These numbers are $-3$ and $2$.

Thus, we can factor:

$t^2 - t - 6 = (t - 3)(t + 2)$

Therefore, the factored form of the quadratic is:

$-t^2 + t + 6 = -(t - 3)(t + 2)$

3. Final factored form:

Substitute this back into the expression for $d(t)$:

$d(t) = 3t \cdot -(t - 3)(t + 2)$

Simplifying the signs:

$d(t) = -3t(t - 3)(t + 2)$

This is the fully factored form of the time function.

Would you like to go over more details or have any further questions?

Related Questions:

1. What is the physical meaning of the roots of the factored function $d(t) = -3t(t - 3)(t + 2)$?
2. How would you find the times when the ship is at the harbour?
3. What is the ship's velocity at $t = 2$ hours using the derivative of the distance function?
4. How can we interpret the sign of the velocity in terms of the ship's movement?
5. What happens to the ship's position after $t = 3$ hours?

Tip:

To determine when the ship changes direction, take the derivative of $d(t)$, find the critical points, and analyze the sign changes!