b Describe the limiting behaviour of the polynomial. c Draw a sketch of the polynomial by finding the roots and the $y$-intercept Question 4 (7 marks) The polynomial $P(x)=x^{3}-2 x^{2}+p x+24$ has roots $\alpha, \beta, \gamma$. a Find the values of $\alpha+\beta+\gamma, \alpha \gamma+\beta \gamma+\alpha \beta$ and $\alpha \beta \gamma$ $b$ If two of the roots are equal in magnitude but opposite in sign, find the third root and hence the value of $p$. c Hence, find: i $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$ ii $\alpha^{2}+\beta^{2}+\gamma^{2} \quad$ iii $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}} \quad$ iv $\frac{1}{\alpha^{2} \beta^{2}}+\frac{1}{\alpha^{2} \gamma^{2}}+\frac{1}{\beta^{2} \gamma^{2}}$ 3

Step 1 ：Using Vieta's formulas, we have \(\alpha + \beta + \gamma = 2\), \(\alpha\beta + \alpha\gamma + \beta\gamma = p\), and \(\alpha\beta\gamma = -24\).

Step 2 ：Since two of the roots are equal in magnitude but opposite in sign, let \(\alpha = -\beta\). Then, \(\gamma = 2 - \alpha - \beta = 2 - \alpha + \alpha = 2\).

Step 3 ：Substituting \(\gamma = 2\) into \(\alpha\beta\gamma = -24\), we get \(\alpha\beta(2) = -24\), so \(\alpha\beta = -12\).

Step 4 ：Since \(\alpha = -\beta\), we have \(\alpha^2 = 12\), so \(\alpha = \sqrt{12}\) and \(\beta = -\sqrt{12}\).

Step 5 ：Substituting \(\alpha = \sqrt{12}\), \(\beta = -\sqrt{12}\), and \(\gamma = 2\) into \(\alpha\beta + \alpha\gamma + \beta\gamma = p\), we get \(p = -12 + 2\sqrt{12} + 2(-\sqrt{12}) = -12\).

Step 6 ：i) To find \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\), we can use the formula \(\frac{\alpha\beta + \alpha\gamma + \beta\gamma}{\alpha\beta\gamma}\), which gives us \(\frac{-12}{-24} = \boxed{\frac{1}{2}}\).

Step 7 ：ii) To find \(\alpha^2 + \beta^2 + \gamma^2\), we can use the formula \((\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)\), which gives us \((2)^2 - 2(-12) = 4 + 24 = \boxed{28}\).

Step 8 ：iii) To find \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}\), we can use the formula \(\frac{(\alpha\beta + \alpha\gamma + \beta\gamma)^2}{\alpha^2\beta^2\gamma^2}\), which gives us \(\frac{(-12)^2}{(12)(12)(4)} = \frac{144}{576} = \boxed{\frac{1}{4}}\).

Step 9 ：iv) To find \(\frac{1}{\alpha^2\beta^2} + \frac{1}{\alpha^2\gamma^2} + \frac{1}{\beta^2\gamma^2}\), we can use the formula \(\frac{\alpha^2\beta^2\gamma^2}{(\alpha\beta\gamma)^2}\), which gives us \(\frac{(12)(12)(4)}{(-24)^2} = \frac{576}{576} = \boxed{1}\).