SOLUTION: Hi, can you completely factor this problem? 5x^2-4x+12 I tried everything, but nothing seems to work. Thanks.

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Question 379889: Hi, can you completely factor this problem? 5x^2-4x+12 I tried everything, but nothing seems to work. Thanks.
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .


Now multiply the first coefficient by the last term to get .


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


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


Factors of :
1,2,3,4,5,6,10,12,15,20,30,60
-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60


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


These factors pair up and multiply to .
1*60 = 60
2*30 = 60
3*20 = 60
4*15 = 60
5*12 = 60
6*10 = 60
(-1)*(-60) = 60
(-2)*(-30) = 60
(-3)*(-20) = 60
(-4)*(-15) = 60
(-5)*(-12) = 60
(-6)*(-10) = 60

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


First NumberSecond NumberSum
1601+60=61
2302+30=32
3203+20=23
4154+15=19
5125+12=17
6106+10=16
-1-60-1+(-60)=-61
-2-30-2+(-30)=-32
-3-20-3+(-20)=-23
-4-15-4+(-15)=-19
-5-12-5+(-12)=-17
-6-10-6+(-10)=-16



From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.


===============================================================



Answer:


So doesn't factor at all (over the rational numbers).


So is prime.


So it's very possible that you also came to this conclusion, but didn't accept it because most math problems have an answer.


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

Also, feel free to check out my tutoring website

Jim

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