SOLUTION: Factor this problem completely x^8-1

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Question 232374: Factor this problem completely
x^8-1
Found 2 solutions by rapaljer, jsmallt9:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
This problem is called a "Difference of two Squares" since x^8 is actually a perfect square x^4*x^4:

x^8-1
(x^4-1)(x^4+1)

The SUM of two Squares does NOT factor like the Difference of Squares factors, so leave it alone. However, notice that the first factor x^4-1 DOES factor again:
(x^2-1)(x^2+1)(x^4+1)

Once again, you have ANOTHER difference of squares that must be factored. However, do NOT try to factor ANY of those SUMS of squares!

(x-1)(x+1)(x^2+1)(x^4+1) FINAL ANSWER!!

For additional help with this VERY important topic of FACTORING, please visit my own website. Do a "Bing" or "Google" search for my last name "Rapalje". Look for "Rapalje Homepage" near the top of the search list. From my Homepage, look for "Basic, Intermediate, and College Algebra: One Step at a Time." Select "Basic Algebra" and look in "Chapter 2" for several topics on Factoring, especially "Difference of Squares section." These sections are supported by my "MATH IN LIVING COLOR" pages in which I solved problems IN COLOR!

In addition, I have FREE videos posted of me teaching this topic (a few years ago before I retired!). To see the videos, from my Homepage, look for "Rapalje Videos in Living Color". The videos and the non-traditional explanations are all FREE! I'm not selling anything!!

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
There are many factoring techniques. But this problem requires only one of them: the "Difference of squares" pattern, a%5E2+-+b%5E2+=+%28a%2Bb%29%28a-b%29.

We just have to realize what makes a term a perfect square. Numbers which are perfect squares are not hard to recognize. But variable factors which are perfect squares are not as obvious. For example, x%5E8 would not seem to be a perfect square because 8 is not a perfect square. But x%5E8 is a perfect square. In fact any variable with an even exponent is a perfect square! The reason for this is the rule for exponents: a%5E%28x%2Ay%29+=+%28a%5Ex%29%5Ey. A specific case of this rule is: a%5E%28y%2A2%29+=+%28a%5Ey%292. On the left side we have an even exponent (since 2 times a whole number is even). On the right side we have something squared. (Similar logic tells us that x%5E15 is a perfect cube because the exponent is divisible by 3, x%5E20 is a perfect 4th power because the exponent is divisible by 4, etc.)

Now we're ready to factor x%5E8+-+1 as s difference of squares:
x%5E8+-+1+=+%28x%5E4%29%5E2+-+1%5E2+=+%28x%5E4%2B1%29%28x%5E4-1%29
When you reduce fractions, you always keep going until you can't go any further. Factoring is like reducing fractions because you keep going until you can't go any further. And if we look at %28x%5E4%2B1%29%28x%5E4-1%29 and if we have learned what makes terms perfect squares, we'll realize that the second factor is another difference of squares:

... and yet another difference of squares!!

... and finally we're done. (Note: There is no "Sum of squares" pattern so those factors will not factor further.)