SOLUTION: factor completely z^4-81

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Question 93564: factor completely
z^4-81
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

factor completely 

z%5E4-81

Write z%5E4 as %28z%5E2%29%5E2

Write 81 as 9%5E2

%28z%5E2%29%5E2-9%5E2

Rule for factoring the difference of two squares:

A%5E2-B%5E2 factors as %28A-B%29%28A%2BB%29

Here A+=+z%5E2 and B=9

So answer = 

%28z%5E2-9%29%28z%5E2%2B9%29

But wait! That's not all! That first factor
z%5E2-9 is also the difference of two 
perfect squares, which can be seen if we 
write the 9 as 3%5E2

%28z%5E2-3%5E2%29%28z%5E2%2B9%29 

Then use the same rule again on the first
parentheses, this time with A=z and B=3

%28z-3%29%28z%2B3%29%28z%5E2%2B9%29

That's the final answer.  A word of caution. Do not
attempt to factor the last factor z%5E2%2B9.  That
is the SUM of two squares, which doesn't factor. Only
the DIFFERENCE of two squares can be factored.

Edwin