How do you factor completely #256x^4-81y^4#?

Answer 1

#(16x^2 +9x^2)(4x+3)(4x-3)#

This is the difference of squares. Easy to do, but often tricky to recognize if you do not know the square numbers. Clues are: there are 2 terms, with a minus sign, the powers are even and the numbers are squares:

Hence:

#(16x^2 +9x^2)color(red)((16x^2-9x^2)) " difference of squares again"#

=#(16x^2 +9x^2)color(red)((4x+3)(4x-3)#

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Answer 2

Jace Anderson

To factor completely (256x^4 - 81y^4), you can use the difference of squares formula:

[a^2 - b^2 = (a + b)(a - b)]

Applying this formula, we get:

[256x^4 - 81y^4 = (16x^2 + 9y^2)(16x^2 - 9y^2)]

Now, (16x^2 - 9y^2) is a difference of squares, so it can be factored further:

[16x^2 - 9y^2 = (4x + 3y)(4x - 3y)]

Therefore, the complete factorization of (256x^4 - 81y^4) is:

[256x^4 - 81y^4 = (16x^2 + 9y^2)(4x + 3y)(4x - 3y)]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.