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Simplify: `root(4)(32x^11y^15)` divided by `root(4)(2x^3y^2)`

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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted September 17, 2013 at 2:54 AM via web

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Simplify:

`root(4)(32x^11y^15)` divided by `root(4)(2x^3y^2)`

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Zaca | Student, Undergraduate | (Level 1) Salutatorian

Posted September 17, 2013 at 3:03 AM (Answer #1)

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Since both the numerator and the denominator are to the fourth root, another way to write the equation is:

`root(4)((32x^11y^15)/(2x^3y^2))`

We can simplify inside the root by dividing like terms:

`root(4)(16x^8y^13)`  

Then take the fourth root (if possible!) of each part under the root (16, x, and y):

`root(4)16 = 2`

`root(4)(x^8) = x^2`

`root(4)(y^13) = y^3(root(4)y)`

Put them together can you get:

`2x^2y^3root(4)y` 

` `  

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 17, 2013 at 3:06 AM (Answer #2)

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The expression `(root(4)(32x^11*y^15))/(root(4)(2x^3y^2))` has to be simplified.

`(root(4)(32x^11*y^15))/(root(4)(2x^3y^2))`

= `(32x^11*y^15)^(1/4)/(2x^3y^2)^(1/4)`

= `((32x^11*y^15)/(2x^3y^2))^(1/4)`

= `(32/2)^(1/4)*x^((11-3)/4)*y^((15 - 2)/4)`

= `16^(1/4)*x^(8/4)*y^(13/4)`

= `2*x^2*y^(13/4)`

The simplified form of `(root(4)(32x^11*y^15))/(root(4)(2x^3y^2)) = 2*x^2*y^(13/4)`

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