Question 762287
There are 3 roots, and the roots on the very left and very right are double roots (since they only touch the x axis and does not pass through)


So there are really 2+1+2 = 5 roots total (2 of which are repeating, only one is a non-repeating root)


So the lowest possible degree is a 5th degree polynomial.


The roots are: -6, 2, 5


The factors would then be:  x - (-6), x-2, x-5 which turn into:  x-6, x-2, x-5


But remember -6 and 5 are double roots, so the factors are really (x-6)^2, (x-2), (x-5)^2



Put this all together to get (x-6)^2 * (x-2) * (x-5)^2


Then stick a constant k out front to get:  k*(x-6)^2 * (x-2) * (x-5)^2


Because f(0) = 4, we know that  


f(x) = k*(x-6)^2 * (x-2) * (x-5)^2


f(0) = k*(0-6)^2 * (0-2) * (0-5)^2


4 = k*(0-6)^2 * (0-2) * (0-5)^2


Now solve for k