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) About Me  (Show Source):
You can put this solution on YOUR website!
This is not factorable with real numbers.

Answer by richwmiller(17219) About Me  (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 6z%5E2-7z%2B3, we can see that the first coefficient is 6, the second coefficient is -7, and the last term is 3.



Now multiply the first coefficient 6 by the last term 3 to get %286%29%283%29=18.



Now the question is: what two whole numbers multiply to 18 (the previous product) and add to the second coefficient -7?



To find these two numbers, we need to list all of the factors of 18 (the previous product).



Factors of 18:

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 18.

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 -7:



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 -7. So 6z%5E2-7z%2B3 cannot be factored.



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Answer:



So 6%2Az%5E2-7%2Az%2B3 doesn't factor at all (over the rational numbers).



So 6%2Az%5E2-7%2Az%2B3 is prime.