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How do you factor completely #4b^4-1024#?

Answer 1

Nathan Axel

#4(b^2 + 16)(b - 4)(b + 4)#

The first step is to take out the common factor of 4.

#rArr 4(b^4 - 256) #

now #(b^4 - 256)" is a difference of squares " #

since #b^4 = b^2xxb^2" and " 256 = 16xx16#

a difference of squares factors as follows

#color(red)(|bar(ul(color(white)(a/a)color(black)( a^2 - b^2 = (a - b)(a + b))color(white)(a/a)|)))#

here a #= b^2" and b = 16 #

#rArr b^4 - 256 = (b^2 - 16)(b^2 + 16) #

now #b^2 - 16" is also a difference of squares " #

and factors as # (b - 4)(b + 4) #

Putting this together gives

#4b^4 - 1024 = 4(b^2 +16)(b - 4)(b + 4) #

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Answer 2

Julia Adams

To factor completely (4b^4 - 1024), you can first factor out the greatest common factor (GCF), which is (4). After factoring out the GCF, you get (4(b^4 - 256)). Then, you can further factor (b^4 - 256) as a difference of squares, resulting in (4(b^2 + 16)(b^2 - 16)). Finally, you can factor (b^2 - 16) as another difference of squares, giving you the completely factored expression: (4(b^2 + 16)(b + 4)(b - 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.