document.write( "Question 103802: New Problem, Factor: \r
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Algebra.Com's Answer #75461 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 $\"15v%5E2-31v%2B10\"$ , we can see that the first coefficient is $\"15\"$ , the second coefficient is $\"-31\"$ , and the last term is $\"10\"$ .

Now multiply the first coefficient $\"15\"$ by the last term $\"10\"$ to get $\"%2815%29%2810%29=150\"$ .

Now the question is: what two whole numbers multiply to $\"150\"$ (the previous product) and add to the second coefficient $\"-31\"$ ?

To find these two numbers, we need to list all of the factors of $\"150\"$ (the previous product).

Factors of $\"150\"$ :

1,2,3,5,6,10,15,25,30,50,75,150

-1,-2,-3,-5,-6,-10,-15,-25,-30,-50,-75,-150

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

These factors pair up and multiply to $\"150\"$ .

1*150 = 150
2*75 = 150
3*50 = 150
5*30 = 150
6*25 = 150
10*15 = 150
(-1)*(-150) = 150
(-2)*(-75) = 150
(-3)*(-50) = 150
(-5)*(-30) = 150
(-6)*(-25) = 150
(-10)*(-15) = 150

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-31\"$ :

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First Number	Second Number	Sum
1	150	1+150=151
2	75	2+75=77
3	50	3+50=53
5	30	5+30=35
6	25	6+25=31
10	15	10+15=25
-1	-150	-1+(-150)=-151
-2	-75	-2+(-75)=-77
-3	-50	-3+(-50)=-53
-5	-30	-5+(-30)=-35
-6	-25	-6+(-25)=-31
-10	-15	-10+(-15)=-25

From the table, we can see that the two numbers $\"-6\"$ and $\"-25\"$ add to $\"-31\"$ (the middle coefficient).

So the two numbers $\"-6\"$ and $\"-25\"$ both multiply to $\"150\"$ and add to $\"-31\"$

Now replace the middle term $\"-31v\"$ with $\"-6v-25v\"$ . Remember, $\"-6\"$ and $\"-25\"$ add to $\"-31\"$ . So this shows us that $\"-6v-25v=-31v\"$ .

$\"15v%5E2%2Bhighlight%28-6v-25v%29%2B10\"$ Replace the second term $\"-31v\"$ with $\"-6v-25v\"$ .

$\"%2815v%5E2-6v%29%2B%28-25v%2B10%29\"$ Group the terms into two pairs.

$\"3v%285v-2%29%2B%28-25v%2B10%29\"$ Factor out the GCF $\"3v\"$ from the first group.

$\"3v%285v-2%29-5%285v-2%29\"$ Factor out $\"5\"$ 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.

$\"%283v-5%29%285v-2%29\"$ Combine like terms. Or factor out the common term $\"5v-2\"$

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Answer:

So $\"15%2Av%5E2-31%2Av%2B10\"$ factors to $\"%283v-5%29%285v-2%29\"$ .

In other words, $\"15%2Av%5E2-31%2Av%2B10=%283v-5%29%285v-2%29\"$ .

Note: you can check the answer by expanding $\"%283v-5%29%285v-2%29\"$ to get $\"15%2Av%5E2-31%2Av%2B10\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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