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solve using the principle of zero (7c+3)(4c-12)=0
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I just don't understand how to do this one since it's not as easy as the last one I asked about and I wanted to know where I messed up
(7c+3)(4c-12)=0 ------------------------------ (7c+3)=0 7c+3=0 7c=-3 7c/7=-3/7 c=-3/7 OR (4c-12)=0 (4(c)+4(-3))=0 (4(c-3))=0 (c-3)=0 c-3=0 c=3 ================= c= -3/7 , 3
oh thank you, that makes sense. I really appreciate the help!!
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