SOLUTION: I need to factor the trinomial completely. 4a^3+12a^2+5a

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Question 790717: I need to factor the trinomial completely.
4a^3+12a^2+5a
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

4a%5E3%2B12a%5E2%2B5a Start with the given expression.


a%284a%5E2%2B12a%2B5%29 Factor out the GCF a.


Now let's try to factor the inner expression 4a%5E2%2B12a%2B5


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Looking at the expression 4a%5E2%2B12a%2B5, we can see that the first coefficient is 4, the second coefficient is 12, and the last term is 5.


Now multiply the first coefficient 4 by the last term 5 to get %284%29%285%29=20.


Now the question is: what two whole numbers multiply to 20 (the previous product) and add to the second coefficient 12?


To find these two numbers, we need to list all of the factors of 20 (the previous product).


Factors of 20:
1,2,4,5,10,20
-1,-2,-4,-5,-10,-20


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


These factors pair up and multiply to 20.
1*20 = 20
2*10 = 20
4*5 = 20
(-1)*(-20) = 20
(-2)*(-10) = 20
(-4)*(-5) = 20

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 12:


First NumberSecond NumberSum
1201+20=21
2102+10=12
454+5=9
-1-20-1+(-20)=-21
-2-10-2+(-10)=-12
-4-5-4+(-5)=-9



From the table, we can see that the two numbers 2 and 10 add to 12 (the middle coefficient).


So the two numbers 2 and 10 both multiply to 20 and add to 12


Now replace the middle term 12a with 2a%2B10a. Remember, 2 and 10 add to 12. So this shows us that 2a%2B10a=12a.


4a%5E2%2Bhighlight%282a%2B10a%29%2B5 Replace the second term 12a with 2a%2B10a.


%284a%5E2%2B2a%29%2B%2810a%2B5%29 Group the terms into two pairs.


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


2a%282a%2B1%29%2B5%282a%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.


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


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So a%284a%5E2%2B12a%2B5%29 then factors further to a%282a%2B5%29%282a%2B1%29


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Answer:


So 4a%5E3%2B12a%5E2%2B5a completely factors to a%282a%2B5%29%282a%2B1%29.


In other words, 4a%5E3%2B12a%5E2%2B5a=a%282a%2B5%29%282a%2B1%29.


Note: you can check the answer by expanding a%282a%2B5%29%282a%2B1%29 to get 4a%5E3%2B12a%5E2%2B5a or by graphing the original expression and the answer (the two graphs should be identical).