document.write( "Question 103800: Level 1, General Polynomials, Degree 2 \r
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\n" ); document.write( "\n" ); document.write( "15 s2 - 19 s + 6
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Algebra.Com's Answer #75459 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"15s%5E2-19s%2B6\"$ , we can see that the first coefficient is $\"15\"$ , the second coefficient is $\"-19\"$ , and the last term is $\"6\"$ .

Now multiply the first coefficient $\"15\"$ by the last term $\"6\"$ to get $\"%2815%29%286%29=90\"$ .

Now the question is: what two whole numbers multiply to $\"90\"$ (the previous product) and add to the second coefficient $\"-19\"$ ?

To find these two numbers, we need to list all of the factors of $\"90\"$ (the previous product).

Factors of $\"90\"$ :

1,2,3,5,6,9,10,15,18,30,45,90

-1,-2,-3,-5,-6,-9,-10,-15,-18,-30,-45,-90

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

These factors pair up and multiply to $\"90\"$ .

1*90 = 90
2*45 = 90
3*30 = 90
5*18 = 90
6*15 = 90
9*10 = 90
(-1)*(-90) = 90
(-2)*(-45) = 90
(-3)*(-30) = 90
(-5)*(-18) = 90
(-6)*(-15) = 90
(-9)*(-10) = 90

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-19\"$ :

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First Number	Second Number	Sum
1	90	1+90=91
2	45	2+45=47
3	30	3+30=33
5	18	5+18=23
6	15	6+15=21
9	10	9+10=19
-1	-90	-1+(-90)=-91
-2	-45	-2+(-45)=-47
-3	-30	-3+(-30)=-33
-5	-18	-5+(-18)=-23
-6	-15	-6+(-15)=-21
-9	-10	-9+(-10)=-19

From the table, we can see that the two numbers $\"-9\"$ and $\"-10\"$ add to $\"-19\"$ (the middle coefficient).

So the two numbers $\"-9\"$ and $\"-10\"$ both multiply to $\"90\"$ and add to $\"-19\"$

Now replace the middle term $\"-19s\"$ with $\"-9s-10s\"$ . Remember, $\"-9\"$ and $\"-10\"$ add to $\"-19\"$ . So this shows us that $\"-9s-10s=-19s\"$ .

$\"15s%5E2%2Bhighlight%28-9s-10s%29%2B6\"$ Replace the second term $\"-19s\"$ with $\"-9s-10s\"$ .

$\"%2815s%5E2-9s%29%2B%28-10s%2B6%29\"$ Group the terms into two pairs.

$\"3s%285s-3%29%2B%28-10s%2B6%29\"$ Factor out the GCF $\"3s\"$ from the first group.

$\"3s%285s-3%29-2%285s-3%29\"$ 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.

$\"%283s-2%29%285s-3%29\"$ Combine like terms. Or factor out the common term $\"5s-3\"$

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Answer:

So $\"15%2As%5E2-19%2As%2B6\"$ factors to $\"%283s-2%29%285s-3%29\"$ .

In other words, $\"15%2As%5E2-19%2As%2B6=%283s-2%29%285s-3%29\"$ .

Note: you can check the answer by expanding $\"%283s-2%29%285s-3%29\"$ to get $\"15%2As%5E2-19%2As%2B6\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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