document.write( "Question 440143: Find three consecutive odd integers such that the product of the second and the third one is 99. \n" ); document.write( "

Algebra.Com's Answer #304151 by rwm(914) $\"\"$ $\"About$

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x,x+2 and x+4
\n" ); document.write( "(x+2)*(x+4)=99
\n" ); document.write( "x^2+6x-91 = 0
\n" ); document.write( "(x+13)*(x-7)=0\r
\n" ); document.write( "\n" ); document.write( "x = -13
\n" ); document.write( "x = +7\r
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"x%5E2%2B6x-91\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"6\"$ , and the last term is $\"-91\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-91\"$ to get $\"%281%29%28-91%29=-91\"$ .

Now the question is: what two whole numbers multiply to $\"-91\"$ (the previous product) and add to the second coefficient $\"6\"$ ?

To find these two numbers, we need to list all of the factors of $\"-91\"$ (the previous product).

Factors of $\"-91\"$ :

1,7,13,91

-1,-7,-13,-91

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

These factors pair up and multiply to $\"-91\"$ .

1*(-91) = -91
7*(-13) = -91
(-1)*(91) = -91
(-7)*(13) = -91

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"6\"$ :

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First Number	Second Number	Sum
1	-91	1+(-91)=-90
7	-13	7+(-13)=-6
-1	91	-1+91=90
-7	13	-7+13=6

From the table, we can see that the two numbers $\"-7\"$ and $\"13\"$ add to $\"6\"$ (the middle coefficient).

So the two numbers $\"-7\"$ and $\"13\"$ both multiply to $\"-91\"$ and add to $\"6\"$

Now replace the middle term $\"6x\"$ with $\"-7x%2B13x\"$ . Remember, $\"-7\"$ and $\"13\"$ add to $\"6\"$ . So this shows us that $\"-7x%2B13x=6x\"$ .

$\"x%5E2%2Bhighlight%28-7x%2B13x%29-91\"$ Replace the second term $\"6x\"$ with $\"-7x%2B13x\"$ .

$\"%28x%5E2-7x%29%2B%2813x-91%29\"$ Group the terms into two pairs.

$\"x%28x-7%29%2B%2813x-91%29\"$ Factor out the GCF $\"x\"$ from the first group.

$\"x%28x-7%29%2B13%28x-7%29\"$ Factor out $\"13\"$ 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.

$\"%28x%2B13%29%28x-7%29\"$ Combine like terms. Or factor out the common term $\"x-7\"$

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Answer:

So $\"x%5E2%2B6%2Ax-91\"$ factors to $\"%28x%2B13%29%28x-7%29\"$ .

In other words, $\"x%5E2%2B6%2Ax-91=%28x%2B13%29%28x-7%29\"$ .

Note: you can check the answer by expanding $\"%28x%2B13%29%28x-7%29\"$ to get $\"x%5E2%2B6%2Ax-91\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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