The cubic polynomial $P(x) = x^3 + r x^2 + s x + t$, where $r$, $s$, and $t$ are real numbers, has three real zeros, $1$, $\alpha$, and $-\alpha$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Using Vieta's relations again, we know that the sum of the product of the roots taken two at a time equals $s$. Hence,

$1 \cdot \alpha + 1 \cdot (-\alpha) + \alpha \cdot (-\alpha) = s \implies s = 0 - \alpha^2 = -\alpha^2.$

Furthermore, for the product of all roots,

$1 \cdot \alpha \cdot (-\alpha) = -t \implies -\alpha = -t \implies t = \alpha.$

Thus,

$s + t = -\alpha^2 + \alpha.$

The position of the particle can be modeled by the equation:

$x(t) = 18 \cos\left( \frac{2\pi}{5} t \right)$

where the amplitude is 18 and the motion is periodic with a period of 5 seconds.

The particle reaches its rest position when it is at the amplitude, which is 18. Halfway to the equilibrium position (which is 0) is 9 units away from the rest position. Given the motion is simple harmonic, it will take a quarter of the period to reach this point.

Therefore, the time taken is:

$\frac{5}{4} = 1.25 \text{ seconds}.$

First, differentiate the position function to find the velocity:

$\frac{dx}{dt} = 54t^2 + 54t + 9.$

At $t = -2$:

$v = 54(-2)^2 + 54(-2) + 9 = -6.$

Verify through substitution into the equation, we can check if:

$v^2 = (54(-2)^2 + 54(-2) + 9)^2.$

Using partial fraction decomposition, express ( \frac{1}{x(1 + x)} ) as:

$\frac{1}{x(1+x)} = \frac{A}{x} + \frac{B}{1+x},$

to find constants A and B, integrate both sides and simplify to verify the expression.

From the integral solution, we know:

$\log{\left| \frac{1}{x} \right|} = 3t + c.$

Using the initial conditions to find constant $c$ and rearranging gives:

$x = \frac{1}{e^{3t + c}} = \frac{k}{e^{3t}},$ where $k$ is determined from initial conditions.