SOLUTION: factor the quadratic expression y^2-8y+6

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Question 278475: factor the quadratic expression y^2-8y+6
Answer by jim_thompson5910(35256) About Me  (Show Source):
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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:



First NumberSecond NumberSum
161+6=7
232+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.