document.write( "Question 238426: x^(2n+1)-2x^(n+1)+x \n" ); document.write( "
Algebra.Com's Answer #175179 by jsmallt9(3758)\"\" \"About 
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\"x%5E%282n%2B1%29-2x%5E%28n%2B1%29%2Bx\"
\n" ); document.write( "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.

\n" ); document.write( "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:
\n" ); document.write( "\"x%5E%282n%29%2Ax%5E1%29-2%28x%5En%2Ax%5E1%29%2Bx\"
\n" ); document.write( "\"x%5E%282n%29%2Ax%29-2%28x%5En%2Ax%29%2Bx\"
\n" ); document.write( "We can now see that x is a common factor which we can factor out:
\n" ); document.write( "\"x%28x%5E%282n%29-2x%5En+%2B+1%29\"
\n" ); document.write( "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:
\n" ); document.write( "\"x%28%28x%5En%29%5E2-2x%5En+%2B+1%29\"
\n" ); document.write( "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:
\n" ); document.write( "\"x%28x%5En+-+1%29%5E2\"

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

\n" ); document.write( "So, if the problem was \"x%5E%282n%2B1%29-2x%5E%28n%2B1%29%2Bx+=+0\", two solutions are:
\n" ); document.write( "x = 0 or x = 1\r
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