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D DPXJube New member Joined Feb 9, 2012 Messages 1 Feb 9, 2012 #1 test test Last edited: Oct 16, 2018
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Feb 9, 2012 #2 Hello, DPXJube! \(\displaystyle \text{Factor by grouping: }\:2n - 6m + 5n^2 - 15mn\) Click to expand... Factor "by pairs". The first two terms have \(\displaystyle 2\) in common; factor it out. The last two terms have \(\displaystyle 5n\) in common; factor it out. . . . . \(\displaystyle 2(n-3m) + 5n(n - 3m)\) We have: .\(\displaystyle 2\underbrace{(n-3m)} + 5n\underbrace{(n - 3m)}\) The two groups have \(\displaystyle (n-3m)\) in common; factor it out. And we have: .\(\displaystyle (n-3m)(2 + 5n)\)
Hello, DPXJube! \(\displaystyle \text{Factor by grouping: }\:2n - 6m + 5n^2 - 15mn\) Click to expand... Factor "by pairs". The first two terms have \(\displaystyle 2\) in common; factor it out. The last two terms have \(\displaystyle 5n\) in common; factor it out. . . . . \(\displaystyle 2(n-3m) + 5n(n - 3m)\) We have: .\(\displaystyle 2\underbrace{(n-3m)} + 5n\underbrace{(n - 3m)}\) The two groups have \(\displaystyle (n-3m)\) in common; factor it out. And we have: .\(\displaystyle (n-3m)(2 + 5n)\)
