Sinc 0=0 then the equation becomes x^2-3x=0 factor out x from both terms using revers distributive (ab+ac=a(b+c)) (x)(x-3)=0 if yz=0 then assume that x and/or y=0 so x=0 and x-3=0
So if you subtract the 0 out from the left side, the equation is now . Since a x is common in both of the terms on the left side, you can factor it out, now making the equation look like . When factoring, each of the separate things (x and (x-3)), must equal zero while the other one doesn't matter. So in order for the equation to be 0, the only potential options for x is 0 and 3. Therefore x=0 and x=3 are the two solutions