which expression should you multiply the numerator and denominator of $\frac{sqrt3{3}}{sqrt3{2x}}$ by to…

which expression should you multiply the numerator and denominator of $\frac{sqrt3{3}}{sqrt3{2x}}$ by to rationalize the denominator?\n$sqrt3{x^{2}}$\n$sqrt3{4x}$\n$sqrt3{4x^{2}}$\n$sqrt3{2x}$\ndone

which expression should you multiply the numerator and denominator of $\frac{sqrt3{3}}{sqrt3{2x}}$ by to rationalize the denominator?\n$sqrt3{x^{2}}$\n$sqrt3{4x}$\n$sqrt3{4x^{2}}$\n$sqrt3{2x}$\ndone

Answer

Explanation:

Step1: Recall cube - root rationalization rule

To rationalize the denominator of a fraction with a cube - root $\sqrt[3]{a}$ in the denominator, we multiply by $\sqrt[3]{a^{2}}$ because $\sqrt[3]{a}\times\sqrt[3]{a^{2}}=\sqrt[3]{a^{3}} = a$. Here, the denominator is $\sqrt[3]{2x}$.

Step2: Determine the rationalizing factor

We need to find a number such that when multiplied by $\sqrt[3]{2x}$, the result is a perfect - cube. We know that $(2x)\times(4x^{2}) = 8x^{3}=(2x)^{3}$. So the factor we multiply the numerator and denominator by is $\sqrt[3]{4x^{2}}$.

Answer:

$\sqrt[3]{4x^{2}}$