Question

Given that $\sqrt{2}$ is a zero of the cubic polynomial $6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2}$ , find its other two zeroes.

Answer 1

Step 1: Determine a quadratic equation

Let $p\left(x\right)=6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2}$ and we know that $x-\sqrt{2}$ is one of the factors of $p\left(x\right)$.

Then,

Step 2: Determine other factors

$$\begin{gathered} \Rightarrow \left(x-\sqrt{2}\right)\left(6x^2+7\sqrt{2}x+4\right) \\ \Rightarrow \left(x-\sqrt{2}\right)\left(6x^2+4\sqrt{2}x+3\sqrt{2}x+4\right)\mathbf{}\mathbf{}[\mathrm{Spliting}\mathrm{middle}\mathrm{term}] \\ \Rightarrow \left(x-\sqrt{2}\right)\left[\left(2x\left(3x+2\sqrt{2}\right)\right)+\left(\sqrt{2}\left(3x+2\sqrt{2}\right)\right)\right][\mathrm{Factorising}\mathrm{the}\mathrm{polynomial}] \\ \Rightarrow \left(x-\sqrt{2}\right)\left(2x+\sqrt{2}\right)\left(3x+2\sqrt{2}\right) \end{gathered}$$

Now, When $p\left(x\right)$ is equal to $0$

$\Rightarrow \left(x=\sqrt{2}\right),\left(x=-\frac{\sqrt{2}}{2}\right),\left(x=-\frac{2\sqrt{2}}{3}\right)$

Hence, other two zeroes are $-\frac{\sqrt{2}}{2}$ are $-\frac{2\sqrt{2}}{3}$.