SOLUTION: Hello Tutor, I was wondering if I could get a bit of help. I am suppose to factor this equation completely could you please tell me if I have done this right or wrong. 6x^

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello Tutor, I was wondering if I could get a bit of help. I am suppose to factor this equation completely could you please tell me if I have done this right or wrong. 6x^      Log On


   



Question 96723: Hello Tutor,
I was wondering if I could get a bit of help. I am suppose to factor this equation completely could you please tell me if I have done this right or wrong.
6x^2 + 7x + 2
I got (3x + 1) (2x + 2)
Thank you for your help.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 6x%5E2%2B7x%2B2, we can see that the first coefficient is 6, the second coefficient is 7, and the last term is 2.



Now multiply the first coefficient 6 by the last term 2 to get %286%29%282%29=12.



Now the question is: what two whole numbers multiply to 12 (the previous product) and add to the second coefficient 7?



To find these two numbers, we need to list all of the factors of 12 (the previous product).



Factors of 12:

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12



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



These factors pair up and multiply to 12.

1*12 = 12
2*6 = 12
3*4 = 12
(-1)*(-12) = 12
(-2)*(-6) = 12
(-3)*(-4) = 12


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 7:



First NumberSecond NumberSum
1121+12=13
262+6=8
343+4=7
-1-12-1+(-12)=-13
-2-6-2+(-6)=-8
-3-4-3+(-4)=-7




From the table, we can see that the two numbers 3 and 4 add to 7 (the middle coefficient).



So the two numbers 3 and 4 both multiply to 12 and add to 7



Now replace the middle term 7x with 3x%2B4x. Remember, 3 and 4 add to 7. So this shows us that 3x%2B4x=7x.



6x%5E2%2Bhighlight%283x%2B4x%29%2B2 Replace the second term 7x with 3x%2B4x.



%286x%5E2%2B3x%29%2B%284x%2B2%29 Group the terms into two pairs.



3x%282x%2B1%29%2B%284x%2B2%29 Factor out the GCF 3x from the first group.



3x%282x%2B1%29%2B2%282x%2B1%29 Factor out 2 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.



%283x%2B2%29%282x%2B1%29 Combine like terms. Or factor out the common term 2x%2B1



===============================================================



Answer:



So 6%2Ax%5E2%2B7%2Ax%2B2 factors to %283x%2B2%29%282x%2B1%29.



In other words, 6%2Ax%5E2%2B7%2Ax%2B2=%283x%2B2%29%282x%2B1%29.



Note: you can check the answer by expanding %283x%2B2%29%282x%2B1%29 to get 6%2Ax%5E2%2B7%2Ax%2B2 or by graphing the original expression and the answer (the two graphs should be identical).