Your problem is #12x^3+12x^2+3x# and you are trying to find its factors. Try factoring out 3x: #3x(4x^2+4x+1)# does the trick to decrease the size of the numbers and the powers. Next, you should look to see if the trinomial that is inside the parentheses can be factored further. #3x(2x+1)(2x+1)# breaks the quadratic polynomial down into two linear factors, which is another goal of factoring. Since the 2x + 1 repeats as a factor, we usually write it with an exponent: #3x(2x+1)^2#.
Sometimes, factoring is a way to solve an equation like yours if it was set = 0. Factoring allows you to use the Zero Product Property to find those solutions. Set each factor = 0 and solve: #3x=0# so x = 0 or #(2x+1) = 0# so 2x = -1 and then x = #-1/2#.
Other times, the factoring can help us to graph the function y = #12x^3+12x^2+3x# by again helping to find the zeros or x-intercepts. They would be (0,0) and #(-1/2,0)#. That can be helpful information to start to graph this function!
