SOLUTION: x^(2n+1)-2x^(n+1)+x

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Question 238426: x^(2n+1)-2x^(n+1)+x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E%282n%2B1%29-2x%5E%28n%2B1%29%2Bx
Is there something missing? Without an equals sign this is just an expression. And all you can do with an expression is simplify it or factor it.

Since your expression does not simplify any farther all we can do is factor it. To see how to factor it I am going to use the property of exponents, a%5E%28%28p%2Bq%29%29+=+a%5Ep%2Aa%5Eq to rewrite the first two terms in a way that makes the common factor easier to see:
x%5E%282n%29%2Ax%5E1%29-2%28x%5En%2Ax%5E1%29%2Bx
x%5E%282n%29%2Ax%29-2%28x%5En%2Ax%29%2Bx
We can now see that x is a common factor which we can factor out:
x%28x%5E%282n%29-2x%5En+%2B+1%29
And now we can factor the second factor as a perfect square trinomial. (It may be easier to see this as a perfect square trinomial if we use the property of exponents, a%5E%28%28p%2Aq%29%29+=+%28a%5Ep%29%5Eq%29, to rewrite the first term:
x%28%28x%5En%29%5E2-2x%5En+%2B+1%29
This fits the perfect square trinomial pattern, a%5E2+-2ab+%2B+b%5E2+=+%28a-b%29%28a-b%29+=+%28a-b%29%5E2, where "a" is x%5En and "b" is 1. Now we can use the pattern to factor it into:
x%28x%5En+-+1%29%5E2

If there was an "= 0" missing in the original equation, we would use factoring to solve it. We would now have:
x%28x%5En+-+1%29%5E2+=+0
And this would be true only if
x+=+0 or x%5En+-+1+=+0
Without knowing n we will not be able to find all the solutions to the second equation. But one solution is x = 1.

So, if the problem was x%5E%282n%2B1%29-2x%5E%28n%2B1%29%2Bx+=+0, two solutions are:
x = 0 or x = 1