SOLUTION: Factoring by using trial factors 2y^2+7y+3

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Question 258024: Factoring by using trial factors 2y^2+7y+3
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Here is how to factor!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 2y%5E2%2B7y%2B3, we can see that the first coefficient is 2, the second coefficient is 7, 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 7?



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 7:



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 the two numbers 1 and 6 add to 7 (the middle coefficient).



So the two numbers 1 and 6 both multiply to 6 and add to 7



Now replace the middle term 7y with y%2B6y. Remember, 1 and 6 add to 7. So this shows us that y%2B6y=7y.



2y%5E2%2Bhighlight%28y%2B6y%29%2B3 Replace the second term 7y with y%2B6y.



%282y%5E2%2By%29%2B%286y%2B3%29 Group the terms into two pairs.



y%282y%2B1%29%2B%286y%2B3%29 Factor out the GCF y from the first group.



y%282y%2B1%29%2B3%282y%2B1%29 Factor out 3 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



%28y%2B3%29%282y%2B1%29 Combine like terms. Or factor out the common term 2y%2B1



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Answer:



So 2%2Ay%5E2%2B7%2Ay%2B3 factors to %28y%2B3%29%282y%2B1%29.



In other words, 2%2Ay%5E2%2B7%2Ay%2B3=%28y%2B3%29%282y%2B1%29.



Note: you can check the answer by expanding %28y%2B3%29%282y%2B1%29 to get 2%2Ay%5E2%2B7%2Ay%2B3 or by graphing the original expression and the answer (the two graphs should be identical).