SOLUTION: Factor by using trial factors 2x^2+3x+1

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Question 257951: Factor by using trial factors 2x^2+3x+1
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
2x^2+3x+1=(x+1)(2 x+1)
Here is how to do it!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 2x%5E2%2B3x%2B1, we can see that the first coefficient is 2, the second coefficient is 3, and the last term is 1.



Now multiply the first coefficient 2 by the last term 1 to get %282%29%281%29=2.



Now the question is: what two whole numbers multiply to 2 (the previous product) and add to the second coefficient 3?



To find these two numbers, we need to list all of the factors of 2 (the previous product).



Factors of 2:

1,2

-1,-2



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



These factors pair up and multiply to 2.

1*2 = 2
(-1)*(-2) = 2


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 3:



First NumberSecond NumberSum
121+2=3
-1-2-1+(-2)=-3




From the table, we can see that the two numbers 1 and 2 add to 3 (the middle coefficient).



So the two numbers 1 and 2 both multiply to 2 and add to 3



Now replace the middle term 3x with x%2B2x. Remember, 1 and 2 add to 3. So this shows us that x%2B2x=3x.



2x%5E2%2Bhighlight%28x%2B2x%29%2B1 Replace the second term 3x with x%2B2x.



%282x%5E2%2Bx%29%2B%282x%2B1%29 Group the terms into two pairs.



x%282x%2B1%29%2B%282x%2B1%29 Factor out the GCF x from the first group.



x%282x%2B1%29%2B1%282x%2B1%29 Factor out 1 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.



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



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Answer:



So 2%2Ax%5E2%2B3%2Ax%2B1 factors to %28x%2B1%29%282x%2B1%29.



In other words, 2%2Ax%5E2%2B3%2Ax%2B1=%28x%2B1%29%282x%2B1%29.



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