SOLUTION: Factor completely: 6x^2+33x+15

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Question 369932: Factor completely: 6x^2+33x+15
Answer by jim_thompson5910(35256) About Me  (Show Source):
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6x%5E2%2B33x%2B15 Start with the given expression.


3%282x%5E2%2B11x%2B5%29 Factor out the GCF 3.


Now let's try to factor the inner expression 2x%5E2%2B11x%2B5


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Looking at the expression 2x%5E2%2B11x%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 11x with x%2B10x. Remember, 1 and 10 add to 11. So this shows us that x%2B10x=11x.


2x%5E2%2Bhighlight%28x%2B10x%29%2B5 Replace the second term 11x with x%2B10x.


%282x%5E2%2Bx%29%2B%2810x%2B5%29 Group the terms into two pairs.


x%282x%2B1%29%2B%2810x%2B5%29 Factor out the GCF x from the first group.


x%282x%2B1%29%2B5%282x%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.


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


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So 3%282x%5E2%2B11x%2B5%29 then factors further to 3%28x%2B5%29%282x%2B1%29


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Answer:


So 6x%5E2%2B33x%2B15 completely factors to 3%28x%2B5%29%282x%2B1%29.


In other words, 6x%5E2%2B33x%2B15=3%28x%2B5%29%282x%2B1%29.


Note: you can check the answer by expanding 3%28x%2B5%29%282x%2B1%29 to get 6x%5E2%2B33x%2B15 or by graphing the original expression and the answer (the two graphs should be identical).


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim