document.write( "Question 893361: Help? Im kinda confused on this one:
\n" ); document.write( "Find three consecutive integers such that the product of the first and the third is 20 more than the second.(Application of Quadratic function.)
\n" ); document.write( "I've tried with
\n" ); document.write( " xz=y+20
\n" ); document.write( "or
\n" ); document.write( " (a sub 1)(a sub 3)= (a sub 2)+ 20
\n" ); document.write( "but cant derive any equations for the three variables \n" ); document.write( "

Algebra.Com's Answer #541227 by richwmiller(17219) $\"\"$ $\"About$

You can put this solution on YOUR website!
x^2+2x-21=0
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"x%5E2%2B2x-21\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"2\"$ , and the last term is $\"-21\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-21\"$ to get $\"%281%29%28-21%29=-21\"$ .

Now the question is: what two whole numbers multiply to $\"-21\"$ (the previous product) and add to the second coefficient $\"2\"$ ?

To find these two numbers, we need to list all of the factors of $\"-21\"$ (the previous product).

Factors of $\"-21\"$ :

1,3,7,21

-1,-3,-7,-21

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

These factors pair up and multiply to $\"-21\"$ .

1*(-21) = -21
3*(-7) = -21
(-1)*(21) = -21
(-3)*(7) = -21

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	-21	1+(-21)=-20
3	-7	3+(-7)=-4
-1	21	-1+21=20
-3	7	-3+7=4

From the table, we can see that there are no pairs of numbers which add to $\"2\"$ . So $\"x%5E2%2B2x-21\"$ cannot be factored.

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Answer:

So $\"x%5E2%2B2%2Ax-21\"$ doesn't factor at all (over the rational numbers).

So $\"x%5E2%2B2%2Ax-21\"$ is prime.

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