Question 206554
The last solution is correct, but here's why {{{x^(2n)-2x^n+1}}} factors to {{{(x^n-1)^2}}}...




{{{x^(2n)-2x^n+1}}} Start with the given expression.



{{{(x^n)^2-2x^n+1}}} Rewrite {{{x^(2n)}}} as {{{(x^n)^2}}}. Note: {{{x^(y*z)=(x^(y))^z}}}



Now let {{{y=x^n}}} (a substitution to make things easier)



{{{y^2-2y+1}}} Replace each {{{x^n}}} with 'y'



Looking at the expression {{{y^2-2y+1}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-2}}}, and the last term is {{{1}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{1}}} to get {{{(1)(1)=1}}}.



Now the question is: what two whole numbers multiply to {{{1}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-2}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{1}}} (the previous product).



Factors of {{{1}}}:

1

-1



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{1}}}.

1*1 = 1
(-1)*(-1) = 1


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-2}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>1+1=2</font></td></tr><tr><td  align="center"><font color=red>-1</font></td><td  align="center"><font color=red>-1</font></td><td  align="center"><font color=red>-1+(-1)=-2</font></td></tr></table>



From the table, we can see that the two numbers {{{-1}}} and {{{-1}}} add to {{{-2}}} (the middle coefficient).



So the two numbers {{{-1}}} and {{{-1}}} both multiply to {{{1}}} <font size=4><b>and</b></font> add to {{{-2}}}



Now replace the middle term {{{-2y}}} with {{{-y-y}}}. Remember, {{{-1}}} and {{{-1}}} add to {{{-2}}}. So this shows us that {{{-y-y=-2y}}}.



{{{y^2+highlight(-y-y)+1}}} Replace the second term {{{-2y}}} with {{{-y-y}}}.



{{{(y^2-y)+(-y+1)}}} Group the terms into two pairs.



{{{y(y-1)+(-y+1)}}} Factor out the GCF {{{y}}} from the first group.



{{{y(y-1)-1(y-1)}}} Factor out {{{1}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(y-1)(y-1)}}} Combine like terms. Or factor out the common term {{{y-1}}}



{{{(y-1)^2}}} Condense the terms.



{{{(x^n-1)^2}}} Plug in {{{y=x^n}}}




So {{{x^(2n)-2x^n+1}}} factors to {{{(x^n-1)^2}}}.



In other words, {{{x^(2n)-2x^n+1=(x^n-1)^2}}}.