SOLUTION: Complete the following statement 6a^2-5a+1=(3a-1)( ). I believe I'm supposed to find the missing set but I cannot figure out the equation.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Complete the following statement 6a^2-5a+1=(3a-1)( ). I believe I'm supposed to find the missing set but I cannot figure out the equation.      Log On


   

Question 103933: Complete the following statement 6a^2-5a+1=(3a-1)( ). I believe I'm supposed to find the missing set but I cannot figure out the equation.
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 6a%5E2-5a%2B1, we can see that the first coefficient is 6, the second coefficient is -5, and the last term is 1.



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



Now the question is: what two whole numbers multiply to 6 (the previous product) and add to the second coefficient -5?



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 -5:



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 -2 and -3 add to -5 (the middle coefficient).



So the two numbers -2 and -3 both multiply to 6 and add to -5



Now replace the middle term -5a with -2a-3a. Remember, -2 and -3 add to -5. So this shows us that -2a-3a=-5a.



6a%5E2%2Bhighlight%28-2a-3a%29%2B1 Replace the second term -5a with -2a-3a.



%286a%5E2-2a%29%2B%28-3a%2B1%29 Group the terms into two pairs.



2a%283a-1%29%2B%28-3a%2B1%29 Factor out the GCF 2a from the first group.



2a%283a-1%29-1%283a-1%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.



%282a-1%29%283a-1%29 Combine like terms. Or factor out the common term 3a-1



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



Answer:



So 6%2Aa%5E2-5%2Aa%2B1 factors to %282a-1%29%283a-1%29.



In other words, 6%2Aa%5E2-5%2Aa%2B1=%282a-1%29%283a-1%29.



Note: you can check the answer by expanding %282a-1%29%283a-1%29 to get 6%2Aa%5E2-5%2Aa%2B1 or by graphing the original expression and the answer (the two graphs should be identical).





So the missing expression is 2a-1