SOLUTION: factor each completely: n^2-11n+10

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Question 398464: factor each completely:
n^2-11n+10
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression n%5E2-11n%2B10, we can see that the first coefficient is 1, the second coefficient is -11, and the last term is 10.


Now multiply the first coefficient 1 by the last term 10 to get %281%29%2810%29=10.


Now the question is: what two whole numbers multiply to 10 (the previous product) and add to the second coefficient -11?


To find these two numbers, we need to list all of the factors of 10 (the previous product).


Factors of 10:
1,2,5,10
-1,-2,-5,-10


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


These factors pair up and multiply to 10.
1*10 = 10
2*5 = 10
(-1)*(-10) = 10
(-2)*(-5) = 10

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -11:


First NumberSecond NumberSum
1101+10=11
252+5=7
-1-10-1+(-10)=-11
-2-5-2+(-5)=-7



From the table, we can see that the two numbers -1 and -10 add to -11 (the middle coefficient).


So the two numbers -1 and -10 both multiply to 10 and add to -11


Now replace the middle term -11n with -n-10n. Remember, -1 and -10 add to -11. So this shows us that -n-10n=-11n.


n%5E2%2Bhighlight%28-n-10n%29%2B10 Replace the second term -11n with -n-10n.


%28n%5E2-n%29%2B%28-10n%2B10%29 Group the terms into two pairs.


n%28n-1%29%2B%28-10n%2B10%29 Factor out the GCF n from the first group.


n%28n-1%29-10%28n-1%29 Factor out 10 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.


%28n-10%29%28n-1%29 Combine like terms. Or factor out the common term n-1


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Answer:


So n%5E2-11n%2B10 factors to %28n-10%29%28n-1%29.


In other words, n%5E2-11n%2B10=%28n-10%29%28n-1%29.


Note: you can check the answer by expanding %28n-10%29%28n-1%29 to get n%5E2-11n%2B10 or by graphing the original expression and the answer (the two graphs should be identical).


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim