SOLUTION: Factor 3x^2-19+35

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Question 737611: Factor 3x^2-19+35
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
do you mean 3x^2-19x+35
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 3x%5E2-19x%2B35, we can see that the first coefficient is 3, the second coefficient is -19, and the last term is 35.



Now multiply the first coefficient 3 by the last term 35 to get %283%29%2835%29=105.



Now the question is: what two whole numbers multiply to 105 (the previous product) and add to the second coefficient -19?



To find these two numbers, we need to list all of the factors of 105 (the previous product).



Factors of 105:

1,3,5,7,15,21,35,105

-1,-3,-5,-7,-15,-21,-35,-105



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



These factors pair up and multiply to 105.

1*105 = 105
3*35 = 105
5*21 = 105
7*15 = 105
(-1)*(-105) = 105
(-3)*(-35) = 105
(-5)*(-21) = 105
(-7)*(-15) = 105


Now let's add up each pair of factors to see if one pair adds to the middle coefficient -19:



First NumberSecond NumberSum
11051+105=106
3353+35=38
5215+21=26
7157+15=22
-1-105-1+(-105)=-106
-3-35-3+(-35)=-38
-5-21-5+(-21)=-26
-7-15-7+(-15)=-22




From the table, we can see that there are no pairs of numbers which add to -19. So 3x%5E2-19x%2B35 cannot be factored.



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Answer:



So 3%2Ax%5E2-19%2Ax%2B35 doesn't factor at all (over the rational numbers).



So 3%2Ax%5E2-19%2Ax%2B35 is prime.