document.write( "Question 848184: I am having trouble factoring the following equation: 2a^2-9a+3 \n" ); document.write( "

Algebra.Com's Answer #510922 by richwmiller(17219) $\"\"$ $\"About$

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It isn't an equation. Do you see an equal sign?
\n" ); document.write( "It is a polynomial expression.
\n" ); document.write( "It can't be factored.
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"2a%5E2-9a%2B3\"$ , we can see that the first coefficient is $\"2\"$ , the second coefficient is $\"-9\"$ , and the last term is $\"3\"$ .

Now multiply the first coefficient $\"2\"$ by the last term $\"3\"$ to get $\"%282%29%283%29=6\"$ .

Now the question is: what two whole numbers multiply to $\"6\"$ (the previous product) and add to the second coefficient $\"-9\"$ ?

To find these two numbers, we need to list all of the factors of $\"6\"$ (the previous product).

Factors of $\"6\"$ :

1,2,3,6

-1,-2,-3,-6

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

These factors pair up and multiply to $\"6\"$ .

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

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-9\"$ :

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First Number	Second Number	Sum
1	6	1+6=7
2	3	2+3=5
-1	-6	-1+(-6)=-7
-2	-3	-2+(-3)=-5

From the table, we can see that there are no pairs of numbers which add to $\"-9\"$ . So $\"2a%5E2-9a%2B3\"$ cannot be factored.

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Answer:

So $\"2%2Aa%5E2-9%2Aa%2B3\"$ doesn't factor at all (over the rational numbers).

So $\"2%2Aa%5E2-9%2Aa%2B3\"$ is prime.

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