SOLUTION: State method and factor completely 2p-6q+pq-3q^ The method to be used is the grouping of terms but I am stuck on the factoring part because the 3q^ is not factorable if I'm not

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: State method and factor completely 2p-6q+pq-3q^ The method to be used is the grouping of terms but I am stuck on the factoring part because the 3q^ is not factorable if I'm not      Log On


   

Question 113019: State method and factor completely
2p-6q+pq-3q^
The method to be used is the grouping of terms but I am stuck on the factoring part because the 3q^ is not factorable if I'm not mistaken. Can you assist me on understanding this problem.
Found 2 solutions by jim_thompson5910, SHUgrad05:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
2p-6q%2Bpq-3q%5E2 Start with the given expression

%282p-6q%29%2B%28pq-3q%5E2%29 Group like terms


2%28p-3q%29%2Bq%28p-3q%29 Factor out the GCF of 2 out of the first group. Factor out the GCF of q out of the second group


%282%2Bq%29%28p-3q%29 Since we have a common term of p-3q, we can combine like terms

So 2p-6q%2Bpq-3q%5E2 factors to %282%2Bq%29%28p-3q%29

Answer by SHUgrad05(58) About Me  (Show Source):
You can put this solution on YOUR website!
Should the problem read: 2p-6q+pq-3q^2???
If so, group the first two terms and the last two terms then factor:
(2p-6q)+(pq-3q^2)
2(p-3q)+q(p-3q)
(2+q)(p-3q)