Question

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

Answer 1

Given polynomial is:-

$p\left(x\right)=6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2}$

Since $x=\sqrt{2}$ is a zero of $p\left(x\right)$

$⟹(x-\sqrt{2})$ will be a factor of $p\left(x\right)$

$x-\sqrt{2})6x^3+\sqrt{2}{x}^2-10x-4\sqrt{2}(6x^2+7\sqrt{2}x+46x^3-6\sqrt{2}{x}^2\underset{–}{-+}7\sqrt{2}{x}^2-10x7\sqrt{2}{x}^2-14x\underset{–}{-+}4x-4\sqrt{2}4x-4\sqrt{2}\underset{–}{-+}\underset{–}{0}$

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

Hence two zeroes are $-\frac{2\sqrt{2}}{3}$ and $x=-\frac{1}{\sqrt{2}}$