document.write( "Question 255374: The sum of the squares of two consecutive odd positive integers is 202. Find the integers. \n" ); document.write( "

Algebra.Com's Answer #187644 by richwmiller(17219) $\"\"$ $\"About$

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n^2+(n+2)^2=202\r
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\n" ); document.write( "(2n+11)(n-9)=0\r
\n" ); document.write( "\n" ); document.write( "Be sure to follow how to factor.\r
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

$\"2%2Ax%5E2%2B4%2Ax-198\"$ Start with the given expression.

$\"2%28x%5E2%2B2x-99%29\"$ Factor out the GCF $\"2\"$ .

Now let's try to factor the inner expression $\"x%5E2%2B2x-99\"$

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Looking at the expression $\"x%5E2%2B2x-99\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"2\"$ , and the last term is $\"-99\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-99\"$ to get $\"%281%29%28-99%29=-99\"$ .

Now the question is: what two whole numbers multiply to $\"-99\"$ (the previous product) and add to the second coefficient $\"2\"$ ?

To find these two numbers, we need to list all of the factors of $\"-99\"$ (the previous product).

Factors of $\"-99\"$ :

1,3,9,11,33,99

-1,-3,-9,-11,-33,-99

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

These factors pair up and multiply to $\"-99\"$ .

1*(-99) = -99
3*(-33) = -99
9*(-11) = -99
(-1)*(99) = -99
(-3)*(33) = -99
(-9)*(11) = -99

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	-99	1+(-99)=-98
3	-33	3+(-33)=-30
9	-11	9+(-11)=-2
-1	99	-1+99=98
-3	33	-3+33=30
-9	11	-9+11=2

From the table, we can see that the two numbers $\"-9\"$ and $\"11\"$ add to $\"2\"$ (the middle coefficient).

So the two numbers $\"-9\"$ and $\"11\"$ both multiply to $\"-99\"$ and add to $\"2\"$

Now replace the middle term $\"2x\"$ with $\"-9x%2B11x\"$ . Remember, $\"-9\"$ and $\"11\"$ add to $\"2\"$ . So this shows us that $\"-9x%2B11x=2x\"$ .

$\"x%5E2%2Bhighlight%28-9x%2B11x%29-99\"$ Replace the second term $\"2x\"$ with $\"-9x%2B11x\"$ .

$\"%28x%5E2-9x%29%2B%2811x-99%29\"$ Group the terms into two pairs.

$\"x%28x-9%29%2B%2811x-99%29\"$ Factor out the GCF $\"x\"$ from the first group.

$\"x%28x-9%29%2B11%28x-9%29\"$ Factor out $\"11\"$ from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

$\"%28x%2B11%29%28x-9%29\"$ Combine like terms. Or factor out the common term $\"x-9\"$

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So $\"2%28x%5E2%2B2x-99%29\"$ then factors further to $\"2%28x%2B11%29%28x-9%29\"$

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Answer:

So $\"2%2Ax%5E2%2B4%2Ax-198\"$ completely factors to $\"2%28x%2B11%29%28x-9%29\"$ .

In other words, $\"2%2Ax%5E2%2B4%2Ax-198=2%28x%2B11%29%28x-9%29\"$ .

Note: you can check the answer by expanding $\"2%28x%2B11%29%28x-9%29\"$ to get $\"2%2Ax%5E2%2B4%2Ax-198\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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