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I got this wrong on a math test and I have no idea why. Directions: Solve equation x(3x-2)(2x+3)=0
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so when the equation is set up this way, you just have to take each term and set it equal to zer by itself, so x=0, 3x-2=0, and 2x+3=0, because if any of the terms equals zero, then the equation will be true because anything multiplied by 0 is 0. so x = 0, x=2/3, and x=-3/2. also when ever you have a cubic equation, you know there will be three solutions, although they may sometimes include imaginary solutions.
thanks
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