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(12-2x)x5+2=28-(3x+3)x2
We move all terms to the left:
(12-2x)x5+2-(28-(3x+3)x2)=0
We add all the numbers together, and all the variables
(-2x+12)x5-(28-(3x+3)x2)+2=0
We multiply parentheses
-2x^6+12x^5-(28-(3x+3)x2)+2=0
We calculate terms in parentheses: -(28-(3x+3)x2), so:
28-(3x+3)x2
determiningTheFunctionDomain -(3x+3)x2+28
We multiply parentheses
-3x^3-3x^2+28
We do not support expression: x^3
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