SOLUTION: completely factor 6x^2+21x+15

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Question 751417: completely factor
6x^2+21x+15
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

6x%5E2%2B21x%2B15 Start with the given expression.


3%282x%5E2%2B7x%2B5%29 Factor out the GCF 3.


Now let's try to factor the inner expression 2x%5E2%2B7x%2B5


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Looking at the expression 2x%5E2%2B7x%2B5, we can see that the first coefficient is 2, the second coefficient is 7, 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 7?


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 7:


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 2 and 5 add to 7 (the middle coefficient).


So the two numbers 2 and 5 both multiply to 10 and add to 7


Now replace the middle term 7x with 2x%2B5x. Remember, 2 and 5 add to 7. So this shows us that 2x%2B5x=7x.


2x%5E2%2Bhighlight%282x%2B5x%29%2B5 Replace the second term 7x with 2x%2B5x.


%282x%5E2%2B2x%29%2B%285x%2B5%29 Group the terms into two pairs.


2x%28x%2B1%29%2B%285x%2B5%29 Factor out the GCF 2x from the first group.


2x%28x%2B1%29%2B5%28x%2B1%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.


%282x%2B5%29%28x%2B1%29 Combine like terms. Or factor out the common term x%2B1


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So 3%282x%5E2%2B7x%2B5%29 then factors further to 3%282x%2B5%29%28x%2B1%29


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Answer:


So 6x%5E2%2B21x%2B15 completely factors to 3%282x%2B5%29%28x%2B1%29.


In other words, 6x%5E2%2B21x%2B15=3%282x%2B5%29%28x%2B1%29.


Note: you can check the answer by expanding 3%282x%2B5%29%28x%2B1%29 to get 6x%5E2%2B21x%2B15 or by graphing the original expression and the answer (the two graphs should be identical).