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August 9th, 2015, 11:38 AM   #1
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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.
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August 9th, 2015, 01:11 PM   #2
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$4x^2+9y^2$ is not factorable over the real numbers
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August 9th, 2015, 08:40 PM   #3
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Quote:
Originally Posted by Dalekcaan1963 View Post
I can understand how the answer is reached: square root both sides
Incorrect. Instead, use A² - B² ≡ (A - B)(A + B).
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August 10th, 2015, 10:00 AM   #4
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Quote:
Originally Posted by Dalekcaan1963 View Post
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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