document.write( "Question 98537: Factor the polynomial completely\r
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Algebra.Com's Answer #71692 by jim_thompson5910(35256) $\"\"$ $\"About$

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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 $\"-2\"$ :

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First Number	Second Number	Sum
1	-1	1+(-1)=0
-1	1	-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, $\"2%2Aa%5E2-4%2Aa-2=2%28a%5E2-2a-1%29\"$ .

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