Question 238426
{{{x^(2n+1)-2x^(n+1)+x}}}
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.<br>
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^((p+q)) = a^p*a^q}}} to rewrite the first two terms in a way that makes the common factor easier to see:
{{{x^(2n)*x^1)-2(x^n*x^1)+x}}}
{{{x^(2n)*x)-2(x^n*x)+x}}}
We can now see that x is a common factor which we can factor out:
{{{x(x^(2n)-2x^n + 1)}}}
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^((p*q)) = (a^p)^q)}}}, to rewrite the first term:
{{{x((x^n)^2-2x^n + 1)}}}
This fits the perfect square trinomial pattern, {{{a^2 -2ab + b^2 = (a-b)(a-b) = (a-b)^2}}}, where "a" is {{{x^n}}} and "b" is 1. Now we can use the pattern to factor it into:
{{{x(x^n - 1)^2}}}<br>
If there was an "= 0" missing in the original equation, we would use factoring to solve it. We would now have:
{{{x(x^n - 1)^2 = 0}}}
And this would be true only if
{{{x = 0}}} or {{{x^n - 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.<br>
So, if the problem was {{{x^(2n+1)-2x^(n+1)+x = 0}}}, two solutions are:
x = 0 or x = 1