SOLUTION: Factor completely. m^3 � 9m^2n + 18n^2m

Question 114197: Factor completely. m^3 � 9m^2n + 18n^2m
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

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 )

Now reintroduce the GCF back in.

---------------------------------
Answer:

So factors to
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