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How do you factor #x^8-256#?

Answer 1

Remember the special products: #(A-B)(A+B)=A^2-B^2# and vice versa.

#x^8-256=(x^4)^2-(16)^2=(x^4-16)(x^4+16)#

You can't factor the second term any further, but the first term can be factored according to the same rule: #x^4-16=(x^2)^2-(4)^2=(x^2-4)(x^2+4)#

And again: #x^2-4=(x-2)(x+2)#

If we put all of this together we get:

#x^8-256=(x-2)(x+2)(x^2+4)(x^4+16)#

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Answer 2

Savannah Anderson

To factor ( x^8 - 256 ), you can use the difference of squares formula, which states that ( a^2 - b^2 = (a - b)(a + b) ). Applying this formula, we have:

[ x^8 - 256 = (x^4)^2 - 16^2 ]

Now, this expression is in the form of ( a^2 - b^2 ), where ( a = x^4 ) and ( b = 16 ). Using the difference of squares formula again, we get:

[ (x^4 - 16)(x^4 + 16) ]

Further factoring the first term using the difference of squares again:

[ (x^2 - 4)(x^2 + 4)(x^4 + 16) ]

Finally, factoring ( x^2 - 4 ) and ( x^2 + 4 ) as the difference and sum of squares:

[ (x - 2)(x + 2)(x^2 + 4)(x^4 + 16) ]

So, the factored form of ( x^8 - 256 ) is ( (x - 2)(x + 2)(x^2 + 4)(x^4 + 16) ).

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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.