The parametric equations of a line are given below - HSC - SSCE Mathematics Extension 1 - Question 11 - 2023 - Paper 1

The word 'CONDOBOLIN' consists of 11 letters where:

• C, N, D, O, B, L, I appear once,
• O appears twice,
• N appears twice.

To find the number of arrangements, use the formula for permutations of a multiset:

$\text{Total arrangements} = \frac{n!}{n_1! \times n_2! \times \dots \times n_k!}$

Where:

• $n$ is the total number of letters = 11
• $n_1 = 2$ for O, $n_2 = 2$ for N

Therefore: $\text{Total arrangements} = \frac{11!}{2! \times 2!} = \frac{39916800}{4} = 9979200$

So, there are 9,979,200 different arrangements.

Using the given polynomial $P(x) = x^3 + ax^2 + bx - 12$ and the factor theorem:

1. Since $x + 1$ is a factor, $P(-1) = 0$:

   $(-1)^3 + a(-1)^2 + b(-1) - 12 = 0$
   $-1 + a - b - 12 = 0 \Rightarrow a - b = 13 \quad (1)$

2. The remainder when $P(x)$ is divided by $x - 2$ is -18:

   $P(2) = 2^3 + a(2^2) + b(2) - 12 = -18$
   $8 + 4a + 2b - 12 = -18 \Rightarrow 4a + 2b = -14 \Rightarrow 2a + b = -7 \quad (2)$

3. Solving equations (1) and (2):

   From (1): $b = a - 13$. Substitute into (2):
   $2a + (a - 13) = -7 \Rightarrow 3a - 13 = -7 \Rightarrow 3a = 6 \Rightarrow a = 2.$
   Now find $b$:
   $b = 2 - 13 = -11.$

Thus, $a = 2$ and $b = -11.$