My Math Forum Factorizing 16x^4-81y^4, why is the answer written like this?

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 August 9th, 2015, 12:38 PM #1 Newbie   Joined: Aug 2015 From: UK Posts: 9 Thanks: 0 Factorizing 16x^4-81y^4, why is the answer written like this? I can understand how the answer is reached: square root both sides to make it (4x^2 + 9y^2) (4x^2 - 9y^2) Then factorize the latter brackets to get a final answer of (4x^2 + 9y^2) (2x + 3y) (2x - 3y) Is the reason for not factorizing further with the first set of brackets because the answer would be the same for both brackets as (2x + 3y) (2x + 3y) or (2x - 3y) (2x - 3y)? Why is this done? I want to be extra sure about the rules on this.
 August 9th, 2015, 02:11 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1676 $4x^2+9y^2$ is not factorable over the real numbers
August 9th, 2015, 09:40 PM   #3
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 Originally Posted by Dalekcaan1963 I can understand how the answer is reached: square root both sides
Incorrect. Instead, use A² - B² ≡ (A - B)(A + B).

August 10th, 2015, 11:00 AM   #4
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Quote:
 Originally Posted by Dalekcaan1963 I can understand how the answer is reached: square root both sides to make it . . .
This shows an incomplete understanding of basic principles. If I said find the sum $\displaystyle 64+49$ would you "square root both sides" to get $\displaystyle 8+7$ and say the answer to $\displaystyle 64+49$ is 15?

You can perform the same operation to both sides of a relation ($\displaystyle =,<,$ etc.)

As has been pointed out, what you want in this problem is to apply the well-known factor pattern called difference of squares $\displaystyle x^2-y^2=(x+y)(x-y)$

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