SOLUTION: Factor the polynomial completely 2a^2 - 4a - 2

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Question 98537: Factor the polynomial completely
2a^2 - 4a - 2

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


2%2Aa%5E2-4%2Aa-2 Start with the given expression.



2%28a%5E2-2a-1%29 Factor out the GCF 2.



Now let's try to factor the inner expression a%5E2-2a-1



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Looking at the expression a%5E2-2a-1, we can see that the first coefficient is 1, the second coefficient is -2, and the last term is -1.



Now multiply the first coefficient 1 by the last term -1 to get %281%29%28-1%29=-1.



Now the question is: what two whole numbers multiply to -1 (the previous product) and add to the second coefficient -2?



To find these two numbers, we need to list all of the factors of -1 (the previous product).



Factors of -1:

1

-1



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



These factors pair up and multiply to -1.

1*(-1) = -1
(-1)*(1) = -1


Now let's add up each pair of factors to see if one pair adds to the middle coefficient -2:



First NumberSecond NumberSum
1-11+(-1)=0
-11-1+1=0




From the table, we can see that there are no pairs of numbers which add to -2. So a%5E2-2a-1 cannot be factored.



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Answer:



So 2%2Aa%5E2-4%2Aa-2 simply factors to 2%28a%5E2-2a-1%29



In other words, 2%2Aa%5E2-4%2Aa-2=2%28a%5E2-2a-1%29.