SOLUTION: factoring by using trial factors 6z^2-7z+3.

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Question 258463: factoring by using trial factors 6z^2-7z+3.
Found 2 solutions by Alan3354, richwmiller:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
This is not factorable with real numbers.
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


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 :

1,2,3,6,9,18

-1,-2,-3,-6,-9,-18



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



These factors pair up and multiply to .

1*18 = 18
2*9 = 18
3*6 = 18
(-1)*(-18) = 18
(-2)*(-9) = 18
(-3)*(-6) = 18


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



First NumberSecond NumberSum
1181+18=19
292+9=11
363+6=9
-1-18-1+(-18)=-19
-2-9-2+(-9)=-11
-3-6-3+(-6)=-9




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.
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