Answer:
The simplified expression is:
[tex]54\sqrt{2}\cdot x^3y^4[/tex]
Step-by-step explanation:
We are asked to find the expression:
[tex]3x\sqrt{648x^4y^8}[/tex]
We know that if any quantity under the square root sign is doubled then it comes out of the radical sign as one.
( for example:
[tex]\sqrt{4}=\sqrt{2\times 2}=2[/tex] )
Hence, on prime factorizing we get:
[tex]648x^4y^8=2\times 2\times 2\times 3\times 3\times 3\times 3\times x\times x\times x\times x\times y\times y\times y\times y\times y\times y \times y\times y\\\\\\648x^4y^8=(2\times 2)\times 2\times (3\times 3)\times (3\times 3)\times (x\times x)\times (x\times x)\times (y\times y)\times (y\times y)\times (y\times y) \times (y\times y)\\\\i.e.\\\\\sqrt{648x^4y^8}=2\times 3\times 3\times x\times x\times y\times y\times y\times y\times \sqrt{2}\\\\\\\sqrt{648x^4y^8}=18x^2y^4\sqrt{2}[/tex]
Hence, we get:
[tex]3x\sqrt{648x^4y^8}=3x\times 18x^2y^4\sqrt{2}\\\\i.e.\\\\3x\sqrt{648x^4y^8}=54\sqrt{2}\cdot x^3y^4[/tex]
