SOLUTION: I'm having a hard time understanding the AC method. I have the following problem that has to be solved using the AC method, and I'm lost. Please help. 6m^6n+7m^5n^2+2m^4n^3

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I'm having a hard time understanding the AC method. I have the following problem that has to be solved using the AC method, and I'm lost. Please help. 6m^6n+7m^5n^2+2m^4n^3      Log On


   

Question 107879This question is from textbook
: I'm having a hard time understanding the AC method. I have the following problem that has to be solved using the AC method, and I'm lost. Please help.
6m^6n+7m^5n^2+2m^4n^3 This question is from textbook

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Take the first coefficient (6) and the last coefficient (2) and multiply them to get 12. Now what two numbers multiply to 12 but add to the middle coefficient of 7?

It turns out that 4 and 3 multiply to 12 and add to 7.

6m%5E6n%2B4m%5E5n%5E2%2B3m%5E5n%5E2%2B2m%5E4n%5E3 Now replace the middle term 7m%5E5n%5E2 with the sum 4m%5E5n%5E2%2B3m%5E5n%5E2 (these two are equivalent)

%286m%5E6n%2B4m%5E5n%5E2%29%2B%283m%5E5n%5E2%2B2m%5E4n%5E3%29 Group like terms


2m%5E5n%283m%2B2n%29%2B%283m%5E5n%5E2%2B2m%5E4n%5E3%29 Factor out the GCF 2m%5E5n out of the first group


2m%5E5n%283m%2B2n%29%2Bm%5E4n%5E2%283m%2B2n%29 Factor out the GCF m%5E4n%5E2 out of the second group


Notice we have a common term of 3m+2n, this means we can combine like terms


%282m%5E5n%2Bm%5E4n%5E2%29%283m%2B2n%29 Combine like terms


Notice how %282m%5E5n%2Bm%5E4n%5E2%29%283m%2B2n%29 foils to 6m%5E6n%2B4m%5E5n%5E2%2B3m%5E5n%5E2%2B2m%5E4n%5E3, so this verifies our answer.