SOLUTION: Please help me factor this equation: {{{x^(2n+1)-2x^(n+1)+x}}} Thank you.

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Question 206554This question is from textbook Introductory Algebra
: Please help me factor this equation: x%5E%282n%2B1%29-2x%5E%28n%2B1%29%2Bx
Thank you.
This question is from textbook Introductory Algebra

Found 3 solutions by Alan3354, jim_thompson5910, stanbon:
Answer by Alan3354(69443) About Me  (Show Source):
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The last solution is correct, but here's why x%5E%282n%29-2x%5En%2B1 factors to %28x%5En-1%29%5E2...



x%5E%282n%29-2x%5En%2B1 Start with the given expression.


%28x%5En%29%5E2-2x%5En%2B1 Rewrite x%5E%282n%29 as %28x%5En%29%5E2. Note: x%5E%28y%2Az%29=%28x%5E%28y%29%29%5Ez


Now let y=x%5En (a substitution to make things easier)


y%5E2-2y%2B1 Replace each x%5En with 'y'


Looking at the expression y%5E2-2y%2B1, 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 %281%29%281%29=1.


Now the question is: what two whole numbers multiply to 1 (the previous product) and add to the second coefficient -2?


To find these two numbers, we need to list all 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:


First NumberSecond NumberSum
111+1=2
-1-1-1+(-1)=-2



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 and 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%5E2%2Bhighlight%28-y-y%29%2B1 Replace the second term -2y with -y-y.


%28y%5E2-y%29%2B%28-y%2B1%29 Group the terms into two pairs.


y%28y-1%29%2B%28-y%2B1%29 Factor out the GCF y from the first group.


y%28y-1%29-1%28y-1%29 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.


%28y-1%29%28y-1%29 Combine like terms. Or factor out the common term y-1


%28y-1%29%5E2 Condense the terms.


%28x%5En-1%29%5E2 Plug in y=x%5En



So x%5E%282n%29-2x%5En%2B1 factors to %28x%5En-1%29%5E2.


In other words, x%5E%282n%29-2x%5En%2B1=%28x%5En-1%29%5E2.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^(2n+1)-2x^(n+1)+x
Factor out "x" to get:
x[(x^n)^2 - 2 x^n +1]
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Let x^n = k
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Substitute to get:
x[k^2 -2k + 1]
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= x(k-1)^2
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Substitute to get:
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= x((x^n)-1)^2
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Cheers,
Stan H.