Question 120534: Factor completely: m^3-9m^2n+18n^2m
If anyone could help I would appricate it, I came up with m(m^2-9mn+18n^2) but it doesnt look right for some reason.
Found 2 solutions by jim_thompson5910, josmiceli:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
 Start with the given expression
 Factor out the GCF 
Now let's focus on the inner expression 
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Looking at we can see that the first term is  and the last term is  where the coefficients are 1 and 18 respectively.
Now multiply the first coefficient 1 and the last coefficient 18 to get 18. Now what two numbers multiply to 18 and add to the middle coefficient -9? Let's list all of the factors of 18:
Factors of 18:
1,2,3,6,9,18
-1,-2,-3,-6,-9,-18 ...List the negative factors as well. This will allow us to find all possible combinations
These factors pair up and multiply to 18
1*18
2*9
3*6
(-1)*(-18)
(-2)*(-9)
(-3)*(-6)
note: remember two negative numbers multiplied together make a positive number
Now which of these pairs add to -9? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -9
| 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 |
From this list we can see that -3 and -6 add up to -9 and multiply to 18
Now looking at the expression , replace  with  (notice adds up to . So it is equivalent to )
Now let's factor  by grouping:
 Group like terms
 Factor out the GCF of out of the first group. Factor out the GCF of  out of the second group
 Since we have a common term of  , we can combine like terms
So factors to
So this also means that factors to (since is equivalent to )
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So our expression goes from and factors further to 
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Answer:
So factors to
Answer by josmiceli(19441)  (Show Source):
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