document.write( "Question 101318: factor completely:
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Algebra.Com's Answer #73783 by jim_thompson5910(35256) $\"\"$ $\"About$

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$\"x%5E2%2B9x=8\"$ Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2B9x-8=0\"$ Subtract 8 from both sides\r
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"x%5E2%2B9x-8\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"9\"$ , and the last term is $\"-8\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-8\"$ to get $\"%281%29%28-8%29=-8\"$ .

Now the question is: what two whole numbers multiply to $\"-8\"$ (the previous product) and add to the second coefficient $\"9\"$ ?

To find these two numbers, we need to list all of the factors of $\"-8\"$ (the previous product).

Factors of $\"-8\"$ :

1,2,4,8

-1,-2,-4,-8

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

These factors pair up and multiply to $\"-8\"$ .

1*(-8) = -8
2*(-4) = -8
(-1)*(8) = -8
(-2)*(4) = -8

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"9\"$ :

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First Number	Second Number	Sum
1	-8	1+(-8)=-7
2	-4	2+(-4)=-2
-1	8	-1+8=7
-2	4	-2+4=2

From the table, we can see that there are no pairs of numbers which add to $\"9\"$ . So $\"x%5E2%2B9x-8\"$ cannot be factored.

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Answer:

So $\"x%5E2%2B9%2Ax-8\"$ doesn't factor at all (over the rational numbers).

So $\"x%5E2%2B9%2Ax-8\"$ is prime.

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