SOLUTION: Factor completely. x^3y + 4x^2y^2 - 21xy^3

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Question 201040: Factor completely.
x^3y + 4x^2y^2 - 21xy^3

Answer by jim_thompson5910(13794) About Me  (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 -21 respectively.

Now multiply the first coefficient 1 and the last coefficient -21 to get -21. Now what two numbers multiply to -21 and add to the middle coefficient 4? Let's list all of the factors of -21:



Factors of -21:
1,3,7,21

-1,-3,-7,-21 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to -21
(1)*(-21)
(3)*(-7)
(-1)*(21)
(-3)*(7)

note: remember, the product of a negative and a positive number is a negative number


Now which of these pairs add to 4? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 4

First NumberSecond NumberSum
1-211+(-21)=-20
3-73+(-7)=-4
-121-1+21=20
-37-3+7=4



From this list we can see that -3 and 7 add up to 4 and multiply to -21


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 completely factors to