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factor 2xy-4x+3y-6
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I would group them first using parenthesis. (2xy-4x)+(3y-6)
For the first parenthesis, factor out 2x. 2xy-4x=2x(y-2) And for the second, factor out 3: 3y-6=3(y-2)
So our solution so far should look something like this... (2xy-4x)+(3y-6) = 2x(y-2) + 3(y-2)
Finally, since we have a common factor of (y-2) that means we need to do one more factorization. (2xy-4x)+(3y-6) = 2x(y-2) + 3(y-2) = (y-2)(2x+3)
Does that help?
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