SOLUTION: n^2+4n-12

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Question 507225: n^2+4n-12
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression n%5E2%2B4n-12, we can see that the first coefficient is 1, the second coefficient is 4, and the last term is -12.


Now multiply the first coefficient 1 by the last term -12 to get %281%29%28-12%29=-12.


Now the question is: what two whole numbers multiply to -12 (the previous product) and add to the second coefficient 4?


To find these two numbers, we need to list all of the factors of -12 (the previous product).


Factors of -12:
1,2,3,4,6,12
-1,-2,-3,-4,-6,-12


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


These factors pair up and multiply to -12.
1*(-12) = -12
2*(-6) = -12
3*(-4) = -12
(-1)*(12) = -12
(-2)*(6) = -12
(-3)*(4) = -12

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 4:


First NumberSecond NumberSum
1-121+(-12)=-11
2-62+(-6)=-4
3-43+(-4)=-1
-112-1+12=11
-26-2+6=4
-34-3+4=1



From the table, we can see that the two numbers -2 and 6 add to 4 (the middle coefficient).


So the two numbers -2 and 6 both multiply to -12 and add to 4


Now replace the middle term 4n with -2n%2B6n. Remember, -2 and 6 add to 4. So this shows us that -2n%2B6n=4n.


n%5E2%2Bhighlight%28-2n%2B6n%29-12 Replace the second term 4n with -2n%2B6n.


%28n%5E2-2n%29%2B%286n-12%29 Group the terms into two pairs.


n%28n-2%29%2B%286n-12%29 Factor out the GCF n from the first group.


n%28n-2%29%2B6%28n-2%29 Factor out 6 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%2B6%29%28n-2%29 Combine like terms. Or factor out the common term n-2


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Answer:


So n%5E2%2B4n-12 factors to %28n%2B6%29%28n-2%29.


In other words, n%5E2%2B4n-12=%28n%2B6%29%28n-2%29.


Note: you can check the answer by expanding %28n%2B6%29%28n-2%29 to get n%5E2%2B4n-12 or by graphing the original expression and the answer (the two graphs should be identical).

Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at jim_thompson5910@hotmail.com
or you can visit my website here: http://www.freewebs.com/jimthompson5910/home.html

Thanks,

Jim