How does one factor x² - n² + 2n - 1?
I am trying to factor by grouping so I did x² - (n² + 2n - 1)
I am trying to factor the stuff in parentheses first and then after that put the x² - by it and factor the diff of squares.
However, I am stuck. I can't find the numbers that multiply to -1 and add to 2.
The answer is supposed to be x² - (n-1)² and then simplified (x+1-n)(x-n+1). How do I achieve this answer? What am I doing wrong?
More Questions From This User See All
Answers & Comments
Verified answer
You were correct in trying to group. Think of all the types of polynomials you have studied. Do you remember studying "the difference of 2 squares"? So try grouping x² and (-1) and then - n² + 2n .
If you can figure out the minus sign with the last two terms, you will have (-1)(n² -tn). Then you can factor out the "n" This should help you enough to finish up by yourself. Good luck
When you group them together, you can't leave the negative of the n² outside the group like x² - (n² + 2n - 1). Otherwise, when you distribute it, it changes all of the values inside the group: x² - n² - 2n + 1, which is not the same as the original. You have to factor out the negative afterwards.
x² - n² + 2n - 1
x² + (-n² + 2n - 1)
x² - (n² - 2n + 1)
x² - (n - 1)²
(x + (n - 1))(x - (n - 1))
(x + n - 1)(x - n + 1) <===
It's supposed to be simplified to (x + n - 1)(x - n + 1), not (x+1-n)(x-n+1), which is just (x - n + 1)².
X^2-(n^2+2n-1)
now just focus on what is in the brackets as that is all that can be factorised
(n-1)(n-1) so do the check by doind n x n = -n^2, -n x -1= 1n, n x -1=-1 and -1 x -1= 1
now it like X^2 + (-1x(n-1)^2)
really hope this helps
When you put it in parenthesis you forgot to reverse the signs of the second two terms the way you did with the first term. In parenthesis it should be:
x² - (n² - 2n + 1)
From here you can factor the square to make it the first part of the answer you gave. This is a difference of two squares which leads to the second part of the anser you gave.