Algebra.Com's Answer #288290 by MathLover1(20850)  You can put this solution on YOUR website! \r \n" );
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document.write( " ........factor out common \r \n" );
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document.write( " .....let see if we can factor \r \n" );
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document.write( " Solved by pluggable solver: Factoring using the AC method (Factor by Grouping) | \n" );
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Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .
Now multiply the first coefficient by the last term to get .
Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?
To find these two numbers, we need to list all of the factors of (the previous product).
Factors of :
1,2,3,6,9,18
-1,-2,-3,-6,-9,-18
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to .
1*18 = 18 2*9 = 18 3*6 = 18 (-1)*(-18) = 18 (-2)*(-9) = 18 (-3)*(-6) = 18
Now let's add up each pair of factors to see if one pair adds to the middle coefficient :
First Number | Second Number | Sum | 1 | 18 | 1+18=19 | 2 | 9 | 2+9=11 | 3 | 6 | 3+6=9 | -1 | -18 | -1+(-18)=-19 | -2 | -9 | -2+(-9)=-11 | -3 | -6 | -3+(-6)=-9 | \n" );
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From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.
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Answer:
So doesn't factor at all (over the rational numbers).
So is prime.
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document.write( "so, since cannot be factored, all you can do is \n" );
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