For the cubic polynomial \(x^3 - 3x + 2\), use the cubic formula described below to find the root \(1\) (with multiplicity \(2\)) and \(-2\). You do not have to prove the validity of the cubic formula.]
For a cubic polynomial \(x^3 + px + q\), the roots can all be found by the following formula, carried out in the complex number system. Let \(s\) be a square root of \(q^2/4 + p^3/27\), set \(A = -q/2 + s\), and then let \(c\) be a cube root of \(A\). Then \(c - p/3c\) is a root of \(x^3 + px + q\).
Solution
1. Identification of parameters:
1.1. For the cubic polynomial \(x^3 - 3x + 2\), the coefficients are \(p = -3\) and \(q = 2\).
1.2. Compute the square root term \(s\):\[ s = \sqrt{\frac{q^2}{4} + \frac{p^3}{27}} = \sqrt{\frac{2^2}{4} + \frac{(-3)^3}{27}} = \sqrt{1 - 1} = 0. \]
1.3. Compute \(A\):\[ A = -\frac{q}{2} + s = -\frac{2}{2} + 0 = -1. \]
2. The three complex cube roots of \(A = -1\):
2.1. The solutions to \(c^3 = -1 = e^{i\pi}\) in \(\mathbb{C}\) are:- \(c_1 = -1\),- \(c_2 = e^{i\pi/3} = \frac{1 + i\sqrt{3}}{2}\),- \(c_3 = e^{-i\pi/3} = \frac{1 - i\sqrt{3}}{2}\).
3. Computation of roots via \(x = c - \frac{p}{3c} = c + \frac{1}{c}\):
3.1. For \(c_1 = -1\):\[ x_1 = c_1 + \frac{1}{c_1} = -1 + \frac{1}{-1} = -2. \]
3.2. For \(c_2 = \frac{1 + i\sqrt{3}}{2}\):Since \(\frac{1}{c_2} = \overline{c_2} = \frac{1 - i\sqrt{3}}{2}\):\[ x_2 = c_2 + \frac{1}{c_2} = \frac{1 + i\sqrt{3}}{2} + \frac{1 - i\sqrt{3}}{2} = 1. \]
3.3. For \(c_3 = \frac{1 - i\sqrt{3}}{2}\):Since \(\frac{1}{c_3} = \overline{c_3} = \frac{1 + i\sqrt{3}}{2}\):\[ x_3 = c_3 + \frac{1}{c_3} = \frac{1 - i\sqrt{3}}{2} + \frac{1 + i\sqrt{3}}{2} = 1. \]
4. Conclusion:The cubic formula yields the roots \(x = -2, 1, 1\), giving the root \(1\) with multiplicity \(2\) and the root \(-2\), which factors \((x-1)^2(x+2) = x^3 - 3x + 2\). Q.E.D.