SOLUTION: I have been asked to factor completely: 44b^2-108b-72 I tried it 2 ways and got 2 different answers. Which is correct? First try -72+ -108b=44b^2 4(-18= -27b+11b^2) 4((-6

Question 798104: I have been asked to factor completely: 44b^2-108b-72
I tried it 2 ways and got 2 different answers. Which is correct?
First try
-72+ -108b=44b^2
4(-18= -27b+11b^2)
4((-6 + -11b)(3 + -1b))
4(-6 + -11b)(3 + -1b)
Second try
4((b-3)(11b+6))
4(b-3)(11b+6)
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
factor completely: 44b^2-108b-72
I tried it 2 ways and got 2 different answers. Which is correct?
First try
-72+ -108b=44b^2
4(-18= -27b+11b^2)
4((-6 + -11b)(3 + -1b))
4(-6 + -11b)(3 + -1b)
Second try
4((b-3)(11b+6))
4(b-3)(11b+6)
------------------
If you multiply the factors and get the original trinomial, then it's correct.
Second try
4((b-3)(11b+6))
4(b-3)(11b+6)
These are the same, and it's correct.
--------------
4(-6 + -11b)(3 + -1b)
This it also correct, but not written in the usual descending order. The 2 binomials are multiplied by -1.
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