SOLUTION: 2x^2+6x+4

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Question 682271: 2x^2+6x+4
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming you want to factor this



2x%5E2%2B6x%2B4 Start with the given expression.


2%28x%5E2%2B3x%2B2%29 Factor out the GCF 2.


Now let's try to factor the inner expression x%5E2%2B3x%2B2


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Looking at the expression x%5E2%2B3x%2B2, we can see that the first coefficient is 1, the second coefficient is 3, and the last term is 2.


Now multiply the first coefficient 1 by the last term 2 to get %281%29%282%29=2.


Now the question is: what two whole numbers multiply to 2 (the previous product) and add to the second coefficient 3?


To find these two numbers, we need to list all of the factors of 2 (the previous product).


Factors of 2:
1,2
-1,-2


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


These factors pair up and multiply to 2.
1*2 = 2
(-1)*(-2) = 2

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 3:


First NumberSecond NumberSum
121+2=3
-1-2-1+(-2)=-3



From the table, we can see that the two numbers 1 and 2 add to 3 (the middle coefficient).


So the two numbers 1 and 2 both multiply to 2 and add to 3


Now replace the middle term 3x with x%2B2x. Remember, 1 and 2 add to 3. So this shows us that x%2B2x=3x.


x%5E2%2Bhighlight%28x%2B2x%29%2B2 Replace the second term 3x with x%2B2x.


%28x%5E2%2Bx%29%2B%282x%2B2%29 Group the terms into two pairs.


x%28x%2B1%29%2B%282x%2B2%29 Factor out the GCF x from the first group.


x%28x%2B1%29%2B2%28x%2B1%29 Factor out 2 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%2B2%29%28x%2B1%29 Combine like terms. Or factor out the common term x%2B1


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So 2%28x%5E2%2B3x%2B2%29 then factors further to 2%28x%2B2%29%28x%2B1%29


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Answer:


So 2x%5E2%2B6x%2B4 completely factors to 2%28x%2B2%29%28x%2B1%29.


In other words, 2x%5E2%2B6x%2B4=2%28x%2B2%29%28x%2B1%29.


Note: you can check the answer by expanding 2%28x%2B2%29%28x%2B1%29 to get 2x%5E2%2B6x%2B4 or by graphing the original expression and the answer (the two graphs should be identical).