document.write( "Question 103930: Factor completely.\r
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Algebra.Com's Answer #75613 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"y=x-2\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"3y%5E2-3y-6\"$ \r
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

$\"3%2Ay%5E2-3%2Ay-6\"$ Start with the given expression.

$\"3%28y%5E2-y-2%29\"$ Factor out the GCF $\"3\"$ .

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

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

Now multiply the first coefficient $\"1\"$ by the last term $\"-2\"$ to get $\"%281%29%28-2%29=-2\"$ .

Now the question is: what two whole numbers multiply to $\"-2\"$ (the previous product) and add to the second coefficient $\"-1\"$ ?

To find these two numbers, we need to list all of the factors of $\"-2\"$ (the previous product).

Factors of $\"-2\"$ :

1,2

-1,-2

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

These factors pair up and multiply to $\"-2\"$ .

1*(-2) = -2
(-1)*(2) = -2

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-1\"$ :

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First Number	Second Number	Sum
1	-2	1+(-2)=-1
-1	2	-1+2=1

From the table, we can see that the two numbers $\"1\"$ and $\"-2\"$ add to $\"-1\"$ (the middle coefficient).

So the two numbers $\"1\"$ and $\"-2\"$ both multiply to $\"-2\"$ and add to $\"-1\"$

Now replace the middle term $\"-1y\"$ with $\"y-2y\"$ . Remember, $\"1\"$ and $\"-2\"$ add to $\"-1\"$ . So this shows us that $\"y-2y=-1y\"$ .

$\"y%5E2%2Bhighlight%28y-2y%29-2\"$ Replace the second term $\"-1y\"$ with $\"y-2y\"$ .

$\"%28y%5E2%2By%29%2B%28-2y-2%29\"$ Group the terms into two pairs.

$\"y%28y%2B1%29%2B%28-2y-2%29\"$ Factor out the GCF $\"y\"$ from the first group.

$\"y%28y%2B1%29-2%28y%2B1%29\"$ Factor out $\"2\"$ 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.

$\"%28y-2%29%28y%2B1%29\"$ Combine like terms. Or factor out the common term $\"y%2B1\"$

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So $\"3%28y%5E2-y-2%29\"$ then factors further to $\"3%28y-2%29%28y%2B1%29\"$

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Answer:

So $\"3%2Ay%5E2-3%2Ay-6\"$ completely factors to $\"3%28y-2%29%28y%2B1%29\"$ .

In other words, $\"3%2Ay%5E2-3%2Ay-6=3%28y-2%29%28y%2B1%29\"$ .

Note: you can check the answer by expanding $\"3%28y-2%29%28y%2B1%29\"$ to get $\"3%2Ay%5E2-3%2Ay-6\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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\n" ); document.write( "\n" ); document.write( "Since $\"3y%5E2-3y-6\"$ factors to $\"%283y-6%29%28y%2B1%29\"$ we can replace y with $\"y=x-2\"$ to get \r
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\n" ); document.write( "\n" ); document.write( " $\"%283%28x%2B2%29-6%29%28%28x%2B2%29%2B1%29\"$ \r
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