Answer:
D. [tex]3x\sqrt{2x}[/tex]
Step-by-step explanation:
The problem gives on the following equation:
[tex]\sqrt{32x^3}+-\sqrt{16x^3}+4\sqrt{x^3}-2\sqrt{x^3}[/tex]
Alongside the information that ([tex]x\geq0[/tex]).
One must bear in mind that the operation ([tex]\sqrt[/tex]) indicates that one has to find the number that when multiplied by itself will yield the number underneath the radical. The easiest way to find such a number is to factor the term underneath the radical. Rewrite the terms under the radical as the product of prime numbers,
[tex]\sqrt{2*2*2*2*2*x*x*x}-\sqrt{2*2*2*2*x*x*x}+4\sqrt{x*x*x}-\sqrt{2*x*x*x}[/tex]
Now remove the duplicate factors from underneath the radical,
[tex]2*2*x\sqrt{2x}-2*2*x\sqrt{x}+4x\sqrt{x}-2x\sqrt{x}[/tex]
Simplify,
[tex]4x\sqrt{2x}-4x\sqrt{x}+4x\sqrt{x}-x\sqrt{2x}[/tex]
[tex]3x\sqrt{2x}[/tex]
