SOLUTION: completely factor the expression r^2-2r+1 a. prime b. r(r-2)+1 c. (r-1)(r-1) d. (r+1)(r-1) If you could please explain I would really appreciate it

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: completely factor the expression r^2-2r+1 a. prime b. r(r-2)+1 c. (r-1)(r-1) d. (r+1)(r-1) If you could please explain I would really appreciate it      Log On


   

Question 631477: completely factor the expression r^2-2r+1
a. prime
b. r(r-2)+1
c. (r-1)(r-1)
d. (r+1)(r-1)
If you could please explain I would really appreciate it
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression r%5E2-2r%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 -2r with -r-r. Remember, -1 and -1 add to -2. So this shows us that -r-r=-2r.


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


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


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


r%28r-1%29-1%28r-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.


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




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Answer:


So r%5E2-2r%2B1 factors to %28r-1%29%28r-1%29.


In other words, r%5E2-2r%2B1=%28r-1%29%28r-1%29 for all values of 'r'

So the answer is choice c).


Note: you can check the answer by expanding %28r-1%29%5E2 to get r%5E2-2r%2B1 or by graphing the original expression and the answer (the two graphs should be identical).

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