document.write( "Question 278475: factor the quadratic expression y^2-8y+6 \n" ); document.write( "

Algebra.Com's Answer #202641 by jim_thompson5910(35256) $\"\"$ $\"About$

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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"y%5E2-8y%2B6\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-8\"$ , and the last term is $\"6\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"6\"$ to get $\"%281%29%286%29=6\"$ .

Now the question is: what two whole numbers multiply to $\"6\"$ (the previous product) and add to the second coefficient $\"-8\"$ ?

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 $\"-8\"$ :

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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 $\"-8\"$ . So $\"y%5E2-8y%2B6\"$ cannot be factored.

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Answer:

So $\"y%5E2-8%2Ay%2B6\"$ doesn't factor at all (over the rational numbers).

So $\"y%5E2-8%2Ay%2B6\"$ is prime.

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