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Reduce to lowest terms `x^2 - 16` over `x^2 + 8x + 16` 

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schooledmom | Salutatorian

Posted June 25, 2013 at 2:43 PM via web

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Reduce to lowest terms

`x^2 - 16`

over

`x^2 + 8x + 16` 

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ishpiro | Teacher | (Level 1) Associate Educator

Posted June 25, 2013 at 2:48 PM (Answer #1)

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Reducing to lowest terms means dividing both denominator and numerator by the greatest common factor.

`x^2 - 16` over `x^2 + 8x + 16` can be written as a fraction:

`(x^2 - 16)/(x^2 + 8x + 16)`

Factor both numerator and denominator:

`x^2 - 16 = (x-4)(x+ 4)` because this is a difference of two squares

`x^2 + 8x + 16 = (x+2)(x+4)`

Therefore (x+ 4) is a common factor that can be canceled:

`(x^2 - 16)/(x^2 + 8x + 16) = ((x-4)(x+4))/((x+2)(x+4)) = (x-4)/(x+4)`

``

The result is `(x-4)/(x+4)`

 

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