1. The problem statement, all variables and given/known data [tex]y\prime=4x^3+6x^2+2x+1[/tex] 2. Relevant equations 3. The attempt at a solution [tex]4x^3+6x^2+2x+1[/tex] [tex]-1 = 2x(2x^2+3x+1)[/tex] (I moved the +1 over so I don't get 1/x, but I figured this should screw up the equation since it is no longer 0 on the left hand side.) [tex]-1 = 2x(2x+1)(x+1)[/tex] So the "zeroes" are -1/2, -1, and 0, which is actually the correct answers, but why does this factoring still work? Doesn't the left hand side have to be 0? It bothers me when I can get the correct answer but I don't know why... Isn't that why they're called "zeroes"? They original equation is [itex]y = x^4 + 2x^3 + x^2 + x + 1[/itex], where deriving gets me the equation that is stated above. Usually pllugging in an x will solve for slope, but it is already given that the slope is 1. I also used common sense to figure out if x was 0, only the constant 1 would remain and 1 = 1, so 0 is a point for sure.