Question

Given that $\sqrt{2}$ is a zero of the polynimial $6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2}$. Find other two zeroes.

Answer 1

Since, it is given that $\sqrt{2}$ is the zero of the polynomial $6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2}$, therefore, $(x-\sqrt{2})$ is the factor of the given polynomial.

We now divide $6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2}$ by $(x-\sqrt{2})$ as shown in the below image:

From the division, we observe that the quotient is $(6x^2+7\sqrt{2}x+4)$ and the remainder is $0$.

Now, we factorize the quotient $6x^2+7\sqrt{2}x+4$ by equating it to $0$ to find the other zeroes of the given polynomial:

$6x^2+7\sqrt{2}x+4=0\Rightarrow 6x^2+3\sqrt{2}x+4\sqrt{2}x+4=0\Rightarrow 3x(2x+\sqrt{2})+2\sqrt{2}(2x+\sqrt{2})=0\Rightarrow (3x+2\sqrt{2})(2x+\sqrt{2})=0\Rightarrow (3x+2\sqrt{2})=0,(2x+\sqrt{2})=0\Rightarrow 3x=-2\sqrt{2},2x=-\sqrt{2}\Rightarrow x=-\frac{2\sqrt{2}}{3},x=-\frac{\sqrt{2}}{2}\Rightarrow x=-\frac{2\sqrt{2}}{3},x=-\frac{1}{\sqrt{2}}$

Hence, the other two zeroes of $6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2}$ are $-\frac{2\sqrt{2}}{3},-\frac{1}{\sqrt{2}}$.