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How do you factor #8x^4-128#?

Answer 1

Hazel Anglin

It is

#8x^4-128=8*(x^4-16)=8*(x^4-2^4)=8*(x^2-4)*(x^2+4)=8*(x-2)*(x+2)*(x^2+4)#

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Answer 2

Avery Alexander

To factor 8x^4 - 128, you can first factor out the greatest common factor, which is 8. Then, apply the difference of squares formula to factor the resulting expression further. This yields: 8(x^4 - 16). Finally, recognize that x^4 - 16 is a difference of squares, which factors into (x^2 - 4)(x^2 + 4). Therefore, the fully factored form of 8x^4 - 128 is 8(x^2 - 4)(x^2 + 4).

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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.