document.write( "Question 258463: factoring by using trial factors 6z^2-7z+3. \n" ); document.write( "

Algebra.Com's Answer #190292 by richwmiller(17219) $\"\"$ $\"About$

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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 $\"-7\"$ :

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First Number	Second Number	Sum
1	18	1+18=19
2	9	2+9=11
3	6	3+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.

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