document.write( "Question 617823: Factor completely 2x^2 – 20x -50
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Algebra.Com's Answer #388592 by richwmiller(17219) $\"\"$ $\"About$

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2*(x^2-10x-25)
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"x%5E2-10x-25\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-10\"$ , and the last term is $\"-25\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-25\"$ to get $\"%281%29%28-25%29=-25\"$ .

Now the question is: what two whole numbers multiply to $\"-25\"$ (the previous product) and add to the second coefficient $\"-10\"$ ?

To find these two numbers, we need to list all of the factors of $\"-25\"$ (the previous product).

Factors of $\"-25\"$ :

1,5,25

-1,-5,-25

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

These factors pair up and multiply to $\"-25\"$ .

1*(-25) = -25
5*(-5) = -25
(-1)*(25) = -25
(-5)*(5) = -25

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-10\"$ :

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First Number	Second Number	Sum
1	-25	1+(-25)=-24
5	-5	5+(-5)=0
-1	25	-1+25=24
-5	5	-5+5=0

From the table, we can see that there are no pairs of numbers which add to $\"-10\"$ . So $\"x%5E2-10x-25\"$ cannot be factored.

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Answer:

So $\"x%5E2-10%2Ax-25\"$ doesn't factor at all (over the rational numbers).

So $\"x%5E2-10%2Ax-25\"$ is prime.

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