SOLUTION: What are the factors of n^2-7n+10

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Question 195821: What are the factors of n^2-7n+10
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression n%5E2-7n%2B10, we can see that the first coefficient is 1, the second coefficient is -7, 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 -7?


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
2*5
(-1)*(-10)
(-2)*(-5)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -7:


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 -2 and -5 add to -7 (the middle coefficient).


So the two numbers -2 and -5 both multiply to 10 and add to -7


Now replace the middle term -7n with -2n-5n. Remember, -2 and -5 add to -7. So this shows us that -2n-5n=-7n.


n%5E2%2Bhighlight%28-2n-5n%29%2B10 Replace the second term -7n with -2n-5n.


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


n%28n-2%29%2B%28-5n%2B10%29 Factor out the GCF n from the first group.


n%28n-2%29-5%28n-2%29 Factor out 5 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-5%29%28n-2%29 Combine like terms. Or factor out the common term n-2

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Answer:


So n%5E2-7n%2B10 factors to %28n-5%29%28n-2%29.


Note: you can check the answer by FOILing %28n-5%29%28n-2%29 to get n%5E2-7n%2B10 or by graphing the original expression and the answer (the two graphs should be identical).