The cubic equation $x^3 + 2x^2 + 5x - 1 = 0$ has roots $\alpha, \beta$ and $\gamma$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2018 - Paper 1

For a cubic equation of the form $x^3 + px^2 + qx + r = 0$, if the roots are $r_1, r_2, r_3$, then:

• $p = -(r_1 + r_2 + r_3)$
• $q = r_1r_2 + r_2r_3 + r_1r_3$
• $r = -r_1r_2r_3$.

For the new roots, we can use these relationships to derive the new coefficients.

For the roots $-\frac{1}{\alpha}, -\frac{1}{\beta}, -\frac{1}{\gamma}$, we have:

• The sum of the new roots: $-\frac{1}{\alpha} - \frac{1}{\beta} - \frac{1}{\gamma} = -\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}$
• The product of the new roots: $-\frac{1}{\alpha} \cdot -\frac{1}{\beta} \cdot -\frac{1}{\gamma} = \frac{1}{\alpha\beta\gamma}$

From the original polynomial, we have $\alpha + \beta + \gamma = -2$, $\alpha\beta + \beta\gamma + \gamma\alpha = 5$, and $\alpha\beta\gamma = 1$.

By substituting the calculated sums and products into the polynomial form:

1. The coefficient of $x^2$ (new $p$) is: $\frac{5}{1} = 5$
2. The coefficient of $x$ (new $q$) is: $-(-2) = 2$
3. The constant term (new $r$) is: $-\frac{1}{1} = -1$

Thus, the new cubic equation is: $x^3 + 5x^2 + 2x - 1 = 0$

Comparing with the provided options, the correct cubic equation is option D: $x^3 + 5x^2 - 2x + 1 = 0$.