document.write( "Question 91437: factoring a quadratic with leading coefficient greater than one
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Algebra.Com's Answer #66383 by jim_thompson5910(35256) $\"\"$ $\"About$

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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"y%5E2-10y-39\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-10\"$ , and the last term is $\"-39\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-39\"$ to get $\"%281%29%28-39%29=-39\"$ .

Now the question is: what two whole numbers multiply to $\"-39\"$ (the previous product) and add to the second coefficient $\"-10\"$ ?

To find these two numbers, we need to list all of the factors of $\"-39\"$ (the previous product).

Factors of $\"-39\"$ :

1,3,13,39

-1,-3,-13,-39

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

These factors pair up and multiply to $\"-39\"$ .

1*(-39) = -39
3*(-13) = -39
(-1)*(39) = -39
(-3)*(13) = -39

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-10\"$ :

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First Number	Second Number	Sum
1	-39	1+(-39)=-38
3	-13	3+(-13)=-10
-1	39	-1+39=38
-3	13	-3+13=10

From the table, we can see that the two numbers $\"3\"$ and $\"-13\"$ add to $\"-10\"$ (the middle coefficient).

So the two numbers $\"3\"$ and $\"-13\"$ both multiply to $\"-39\"$ and add to $\"-10\"$

Now replace the middle term $\"-10y\"$ with $\"3y-13y\"$ . Remember, $\"3\"$ and $\"-13\"$ add to $\"-10\"$ . So this shows us that $\"3y-13y=-10y\"$ .

$\"y%5E2%2Bhighlight%283y-13y%29-39\"$ Replace the second term $\"-10y\"$ with $\"3y-13y\"$ .

$\"%28y%5E2%2B3y%29%2B%28-13y-39%29\"$ Group the terms into two pairs.

$\"y%28y%2B3%29%2B%28-13y-39%29\"$ Factor out the GCF $\"y\"$ from the first group.

$\"y%28y%2B3%29-13%28y%2B3%29\"$ 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.

$\"%28y-13%29%28y%2B3%29\"$ Combine like terms. Or factor out the common term $\"y%2B3\"$

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Answer:

So $\"y%5E2-10%2Ay-39\"$ factors to $\"%28y-13%29%28y%2B3%29\"$ .

In other words, $\"y%5E2-10%2Ay-39=%28y-13%29%28y%2B3%29\"$ .

Note: you can check the answer by expanding $\"%28y-13%29%28y%2B3%29\"$ to get $\"y%5E2-10%2Ay-39\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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