The equation $4k^3 - 27k + k = 0$ has a double root - HSC - SSCE Mathematics Extension 2 - Question 5 - 2002 - Paper 1

Given that the polynomial is $P(x) = x^3 - 5x^2 + 5$. To find the roots $\alpha - 1, \beta - 1, \gamma - 1$, we substitute:

$Q(x) = P(x + 1) = (x + 1)^3 - 5(x + 1)^2 + 5.$

Expanding this gives:

$Q(x) = x^3 + 3x^2 + 3x + 1 - 5(x^2 + 2x + 1) + 5 = x^3 - 2x + 1.$

Using the previously found polynomial $P(x)$, we can find the roots. The required polynomial can be derived from:

$R(x) = P(\sqrt{x})P(-\sqrt{x}).$

This results in:

$R(x) = (x^3 - 5x^{3/2} + 5)(x^3 + 5x^{3/2} + 5),$

leading us to the required polynomial with integer coefficients after simplification.

Using the property of polynomial roots: $\alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha) + 3\alpha \beta \gamma.$

From the known sums, we have:

To find the equation of the tangent to the ellipse at point $P(x_1, y_1)$, we use the standard form of the tangent line in ellipses. The derived equation is:

$\frac{x_1 x}{a^2} + \frac{y_1 y}{b^2} = 1.$

This utilizes the point-slope form tangent to an ellipse.

To derive the chord of contact from the point $T(x_0, y_0)$ to the ellipse, we apply the standard form given the coordinates. The resulting equation is:

$\frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1,$

which represents the line connecting point $T$ to the ellipse.