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How do you factor #u^4-81#?

Answer 1

Nathan Axel

#u^4-81 = (u-3)(u+3)(u^2+9)#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We can use this a couple of times to derive the factors with Real coefficients, as follows:

#u^4-81 = (u^2)^2-9^2#

#color(white)(u^4-81) = (u^2-9)(u^2+9)#

#color(white)(u^4-81) = (u^2-3^2)(u^2+9)#

#color(white)(u^4-81) = (u-3)(u+3)(u^2+9)#

The remaining quadratic factor has no simpler linear factors with Real coefficients since #u^2+9 >= 9# for any Real value of #u#, hence no Real zeros or corresponding linear factors.

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Answer 2

Ethan Adkins

To factor ( u^4 - 81 ), you can use the difference of squares formula, which states that ( a^2 - b^2 = (a + b)(a - b) ). In this case, ( a = u^2 ) and ( b = 9 ). So, ( u^4 - 81 ) factors into ( (u^2 + 9)(u^2 - 9) ). Then, you can factor ( u^2 - 9 ) further using the difference of squares formula again to get ( (u^2 + 9)(u + 3)(u - 3) ).

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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.