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Solve number 2 and 3. There is something missing in the question number 3, it should be (the length of edge AB is 8 cm greater than BC)

Let's solve both problems step by step.

### Problem 2: Finding \( n \) and Rational Roots
Given the equation:
\[
x^3 - 5x^2 + 8x + n = 0
\]
It is stated that this equation has a pair of twin integer roots.

#### Part (a) - Finding the Value of \( n \)
A twin root means that there are two roots that are equal. Let's denote the twin root by \( r \). Since the equation is cubic, the possible roots will be \( r, r, \) and some other root \( s \). Therefore, we can express the polynomial as:
\[
(x - r)^2(x - s) = 0
\]
Expanding this, we get:
\[
(x^2 - 2rx + r^2)(x - s) = x^3 - sx^2 - 2rx^2 + 2rsx + r^2x - r^2s
\]
Simplifying, this becomes:
\[
x^3 - (s + 2r)x^2 + (2rs + r^2)x - r^2s = 0
\]
Comparing with the given equation \( x^3 - 5x^2 + 8x + n = 0 \), we have:
\[
s + 2r = 5, \quad 2rs + r^2 = 8, \quad \text{and} \quad -r^2s = n
\]
First, solve \( s + 2r = 5 \). This gives us \( s = 5 - 2r \).

Substitute \( s = 5 - 2r \) into \( 2rs + r^2 = 8 \):
\[
2r(5 - 2r) + r^2 = 8
\]
\[
10r - 4r^2 + r^2 = 8
\]
\[
-3r^2 + 10r - 8 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-10 \pm \sqrt{10^2 - 4(-3)(-8)}}{-6} = \frac{-10 \pm \sqrt{100 - 96}}{-6} = \frac{-10 \pm 2}{-6}
\]
This gives:
\[
r = \frac{-8}{-6} = \frac{4}{3} \quad \text{(not an integer, discard this)} \quad \text{or} \quad r = \frac{-12}{-6} = 2
\]
So, \( r = 2 \). Substituting \( r = 2 \) back into \( s = 5 - 2r \), we get \( s = 1 \).

Now, substitute into \( n = -r^2s \):
\[
n = -(2^2)(1) = -4
\]

#### Part (b) - Finding the Rational Roots
The rational roots are \( r = 2 \) and \( s = 1 \).

### Problem 3: Calculating the Total Surface Area of the Cuboid

Given:
- \( AE = x + 4 \) meters.
- Base area \( ABCD = 33 \) square meters.
- \( AB = BC + 8 \).

Let \( AB = a \) and \( BC = b \). The area of \( ABCD \) gives:
\[
a \times b = 33 \quad \text{and} \quad a = b + 8
\]
Substituting \( a = b + 8 \) into \( ab = 33 \):
\[
(b + 8)b = 33 \quad \Rightarrow \quad b^2 + 8b - 33 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
b = \frac{-8 \pm \sqrt{8^2 + 4(1)(33)}}{2} = \frac{-8 \pm \sqrt{64 + 132}}{2} = \frac{-8 \pm \sqrt{196}}{2} = \frac{-8 \pm 14}{2}
\]
So,
\[
b = 3 \quad (\text{since negative length is not possible})
\]
Thus, \( a = 3 + 8 = 11 \).

To calculate the total surface area of the cuboid, note:
- Length \( a = 11 \) m
- Width \( b = 3 \) m
- Height \( h = x + 4 \) m.

The total surface area \( S \) is:
\[
S = 2(ab + bc + ca) = 2(11 \times 3 + 3 \times (x + 4) + (x + 4) \times 11)
\]
Simplify:
\[
S = 2(33 + 3x + 12 + 11x + 44) = 2(89 + 14x) = 178 + 28x \, \text{square meters}
\]

Let me know if you need any more details or have any questions. 

**Relative Questions:**
1. How would the solution change if the cubic equation had non-twin roots?
2. Can you derive the general form of the cubic equation given its roots?
3. How do surface area formulas change if the shape were a cylinder instead of a cuboid?
4. Can you explain why the quadratic formula was used to solve for \( b \) in the cuboid problem?
5. How would you find the volume of the cuboid given its dimensions?

**Tip:** When solving quadratic equations, always check if the discriminant is a perfect square for easier root calculations.

Learn how to solve a cubic equation with twin roots and calculate the total surface area of a cuboid using given dimensions. This problem covers cubic equations, quadratic equations, and surface area calculations. Suitable for high school students.

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