SOLUTION: Factor the trinomial completely x^2+18xy+81y^2

Question 191474: Factor the trinomial completely
x^2+18xy+81y^2
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Looking at we can see that the first term is and the last term is where the coefficients are 1 and 81 respectively.

Now multiply the first coefficient 1 and the last coefficient 81 to get 81. Now what two numbers multiply to 81 and add to the middle coefficient 18? Let's list all of the factors of 81:

Factors of 81:
1,3,9,27

-1,-3,-9,-27 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to 81
1*81
3*27
9*9
(-1)*(-81)
(-3)*(-27)
(-9)*(-9)

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 18? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 18

First Number	Second Number	Sum
1	81	1+81=82
3	27	3+27=30
9	9	9+9=18
-1	-81	-1+(-81)=-82
-3	-27	-3+(-27)=-30
-9	-9	-9+(-9)=-18

From this list we can see that 9 and 9 add up to 18 and multiply to 81

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 )

note: is equivalent to since the term occurs twice. So also factors to

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Answer:
So factors to

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