We know that the zeroes of an equation are #0,0,2,3#. We can also write this information as #x=0; x=0; x=2; x=3#. Now, we can use this information to develop the factors of an equation. We just need the equation to be equal to zero. That won't change the #x=0#s, but #x=2# becomes #x-2=0#, and #x=3# is #x-3=0#.
Now we have #(x)(x)(x-2)(x-3)=0#. We can multiply these factors to find an equation. Well, #(x)*(x)# is #(x^2)#, and #(x-2)*(x-3)# is #(x^2-3x-2x+6)#, or #x^2-5x+6#.
Right now we have #(x^2)*(x^2-5x+6)#, and we can simplify that to #x^4-5x^3+6x^2#. So now we have our equation. Now, there could be many more equations than this one that will give you zeroes as #x=0,0,2,3#, but this is one of the possible polynomials. One fast way to check if your answer is reasonable is to look at the highest power (in our problem, it's #4#, from #x^color(red)(4)-5x^3+6x^2#). That number should match the number of zeroes we started out with. In our case it does, so that's one check confirming we did our math correctly. Another one is to graph our equation and see if our zeroes are the same ones we started with.
graph{x^4-5x^3+6x^2=y}
And we do! You notice that we have a parabola at #x=0#. Well, we have two #x#s, or #x^2#, which is the equation for a parabola. So that also looks right.
