Given that \(\sqrt{2}\) is a zero of \(6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}\), find the other two zeroes.
Other zeroes: \(x = -\dfrac{\sqrt{2}}{2}\), \(x = -\dfrac{2\sqrt{2}}{3}\).
Step 1: Use factor theorem.
Since \(\sqrt{2}\) is a zero, \(x - \sqrt{2}\) is a factor.
Divide \(6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}\) by \(x - \sqrt{2}\).
Step 2: Do the division (or synthetic division).
Quotient obtained: \(6x^2 + 7\sqrt{2}x + 4\).
(You can verify by multiplication.)
Step 3: Solve the quadratic factor.
\(6x^2 + 7\sqrt{2}x + 4 = 0\).
Discriminant: \((7\sqrt{2})^2 - 4\cdot 6 \cdot 4 = 98 - 96 = 2\).
Roots:
\(x = \dfrac{-7\sqrt{2} \pm \sqrt{2}}{12} = \dfrac{-6\sqrt{2}}{12},\,\dfrac{-8\sqrt{2}}{12}\)
\(= -\dfrac{\sqrt{2}}{2},\,-\dfrac{2\sqrt{2}}{3}\).