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Factor completely: 10xy + 3y + 20ax + 6a (10x − 3)(y − 2a) (10x − 3)(y + 2a) (10x + 3)(y + 2a) (10x + 3)(y − 2a)
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Group the terms in pairs.
the third choice....
cuz first you factor out (10x + 3) from each side to get \[(10x+3)(y+y) + (10x+3)(2a+2a)\]
facotr out the (10x+3) from the whole equation so you get... \[(10x+3)(2y+4a)\] in which the closest answer that you have would be the third chioce... but yeahhh its wrong...
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