SOLUTION: some help please? factor completely 6c^2-33c+15

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Question 270249: some help please?
factor completely
6c^2-33c+15
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First factor out the GCF 3 to get 3%282c%5E2-11c%2B5%29


Now let's factor 2c%5E2-11c%2B5


Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 2c%5E2-11c%2B5, we can see that the first coefficient is 2, the second coefficient is -11, and the last term is 5.



Now multiply the first coefficient 2 by the last term 5 to get %282%29%285%29=10.



Now the question is: what two whole numbers multiply to 10 (the previous product) and add to the second coefficient -11?



To find these two numbers, we need to list all of the factors of 10 (the previous product).



Factors of 10:

1,2,5,10

-1,-2,-5,-10



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



These factors pair up and multiply to 10.

1*10 = 10
2*5 = 10
(-1)*(-10) = 10
(-2)*(-5) = 10


Now let's add up each pair of factors to see if one pair adds to the middle coefficient -11:



First NumberSecond NumberSum
1101+10=11
252+5=7
-1-10-1+(-10)=-11
-2-5-2+(-5)=-7




From the table, we can see that the two numbers -1 and -10 add to -11 (the middle coefficient).



So the two numbers -1 and -10 both multiply to 10 and add to -11



Now replace the middle term -11c with -c-10c. Remember, -1 and -10 add to -11. So this shows us that -c-10c=-11c.



2c%5E2%2Bhighlight%28-c-10c%29%2B5 Replace the second term -11c with -c-10c.



%282c%5E2-c%29%2B%28-10c%2B5%29 Group the terms into two pairs.



c%282c-1%29%2B%28-10c%2B5%29 Factor out the GCF c from the first group.



c%282c-1%29-5%282c-1%29 Factor out 5 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.



%28c-5%29%282c-1%29 Combine like terms. Or factor out the common term 2c-1



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Answer:



So 2%2Ac%5E2-11%2Ac%2B5 factors to %28c-5%29%282c-1%29.



In other words, 2%2Ac%5E2-11%2Ac%2B5=%28c-5%29%282c-1%29.



Note: you can check the answer by expanding %28c-5%29%282c-1%29 to get 2%2Ac%5E2-11%2Ac%2B5 or by graphing the original expression and the answer (the two graphs should be identical).





So 6c%5E2-33c%2B15 completely factors to 3%28c-5%29%282c-1%29