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

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor the trinomial completely x^2+18xy+81y^2      Log On


   



Question 191474: Factor the trinomial completely
x^2+18xy+81y^2

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at x%5E2%2B18xy%2B81y%5E2 we can see that the first term is x%5E2 and the last term is 81y%5E2 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 NumberSecond NumberSum
1811+81=82
3273+27=30
999+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 x%5E2%2B18xy%2B81y%5E2, replace 18xy with 9xy%2B9xy (notice 9xy%2B9xy adds up to 18xy. So it is equivalent to 18xy)

x%5E2%2Bhighlight%289xy%2B9xy%29%2B81y%5E2


Now let's factor x%5E2%2B9xy%2B9xy%2B81y%5E2 by grouping:


%28x%5E2%2B9xy%29%2B%289xy%2B81y%5E2%29 Group like terms


x%28x%2B9y%29%2B9y%28x%2B9y%29 Factor out the GCF of x out of the first group. Factor out the GCF of 9y out of the second group


%28x%2B9y%29%28x%2B9y%29 Since we have a common term of x%2B9y, we can combine like terms

So x%5E2%2B9xy%2B9xy%2B81y%5E2 factors to %28x%2B9y%29%28x%2B9y%29


So this also means that x%5E2%2B18xy%2B81y%5E2 factors to %28x%2B9y%29%28x%2B9y%29 (since x%5E2%2B18xy%2B81y%5E2 is equivalent to x%5E2%2B9xy%2B9xy%2B81y%5E2)


note: %28x%2B9y%29%28x%2B9y%29 is equivalent to %28x%2B9y%29%5E2 since the term x%2B9y occurs twice. So x%5E2%2B18xy%2B81y%5E2 also factors to %28x%2B9y%29%5E2



------------------------------------------------------------



Answer:
So x%5E2%2B18xy%2B81y%5E2 factors to %28x%2B9y%29%5E2