To solve the problem of finding the dimensions of the rectangular solid, we need to use the given volume and the relationships between the dimensions in terms of $x$.

1. Identify the dimensions in terms of $x$:

   • Length: $x + 3$
   • Width: $x - 1$
   • Height: $x - 2$

2. Volume formula: The volume $V$ of a rectangular solid is given by: $V = \text{Length} \times \text{Width} \times \text{Height}$ So we have: $V = (x + 3)(x - 1)(x - 2)$

3. Given volume: The problem states that the volume $V$ is a positive even integer between 39 and 45. The possible values are 40, 42, and 44.

4. Determine the correct $x$: We need to solve for $x$ by trying each possible volume.

   • For $V = 40$: $(x + 3)(x - 1)(x - 2) = 40$
   • For $V = 42$: $(x + 3)(x - 1)(x - 2) = 42$
   • For $V = 44$: $(x + 3)(x - 1)(x - 2) = 44$

Let's solve these equations one by one.

Solving for $V = 40$:

$(x + 3)(x - 1)(x - 2) = 40$ Expanding the left side: $(x + 3)(x^2 - 3x + 2) = 40$ $x^3 - 3x^2 + 2x + 3x^2 - 9x + 6 = 40$ $x^3 - 7x + 6 = 40$ $x^3 - 7x - 34 = 0$

Solving for $V = 42$:

$(x + 3)(x - 1)(x - 2) = 42$ Expanding the left side: $(x + 3)(x^2 - 3x + 2) = 42$ $x^3 - 3x^2 + 2x + 3x^2 - 9x + 6 = 42$ $x^3 - 7x + 6 = 42$ $x^3 - 7x - 36 = 0$

Solving for $V = 44$:

$(x + 3)(x - 1)(x - 2) = 44$ Expanding the left side: $(x + 3)(x^2 - 3x + 2) = 44$ $x^3 - 3x^2 + 2x + 3x^2 - 9x + 6 = 44$ $x^3 - 7x + 6 = 44$ $x^3 - 7x - 38 = 0$

Now we need to find the integer solutions for these cubic equations.

Testing $x^3 - 7x - 34 = 0$:

Using trial and error or a root-finding method:

• $x = 4$: $4^3 - 7(4) - 34 = 64 - 28 - 34 = 2 \quad (\text{not } 0)$
• $x = 5$: $5^3 - 7(5) - 34 = 125 - 35 - 34 = 56 \quad (\text{not } 0)$

No solution found here.

Testing $x^3 - 7x - 36 = 0$:

• $x = 4$: $4^3 - 7(4) - 36 = 64 - 28 - 36 = 0 \quad (\text{solution})$

Thus, $x = 4$ is a solution when $V = 42$.

Testing $x^3 - 7x - 38 = 0$:

Using trial and error:

• $x = 4$: $4^3 - 7(4) - 38 = 64 - 28 - 38 = -2 \quad (\text{not } 0)$

No solution found here.

Conclusion:

For $x = 4$, the volume $V$ is 42, which fits within the given range. The dimensions are:

• Length: $x + 3 = 7$
• Width: $x - 1 = 3$
• Height: $x - 2 = 2$

Thus, the dimensions of the rectangular solid are 7, 3, and 2.

Would you like further details or have any questions?

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Here are 8 related questions to consider:

1. How can we solve cubic equations more efficiently?
2. What is the significance of volume in determining the dimensions of a solid?
3. How do we verify the correctness of our solutions in algebraic problems?
4. Can there be more than one solution for a given volume within a specified range?
5. What are the properties of a rectangular solid in geometry?
6. How does changing one dimension affect the volume of a rectangular solid?
7. What are some real-world applications of finding dimensions of solids?
8. How can we use polynomial equations to solve other types of geometric problems?

Tip: Always check your solutions by substituting them back into the original equation to ensure they satisfy all given conditions.