Question

Please help!!

Answer 1

Answer:   $\text{x}\sqrt{3\text{x}}$   which is Choice A

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Work Shown:

Let $y = \sqrt{3\text{x}}$

So,

$4\text{x}\sqrt{3\text{x}}-\text{x}\sqrt{3\text{x}}-2\text{x}\sqrt{3\text{x}}\\\\4\text{x}y-\text{x}y-2\text{x}y\\\\(4\text{x}-\text{x}-2\text{x})y\\\\\text{x}y\\\\\text{x}\sqrt{3\text{x}}$

This therefore means:

$4\text{x}\sqrt{3\text{x}}-\text{x}\sqrt{3\text{x}}-2\text{x}\sqrt{3\text{x}}=\text{x}\sqrt{3\text{x}}$

as long is x is nonnegative.