Find the critical #s of the fn
f(x)= x^4(x-3)^3

I have gotten to x^3(7x^3-54x^2+135x-108) but do not know how to factor it into what should be x^3(x-3)^2(7x-12)

Wolfram says you should be able to get: 7 x^6 - 54 x^5 + 135 x^4 - 108 x^3 when you expand the derivative of the original function.

You don't need to factor it down. Factoring helps because it gives you free roots, but it isn't a requirement. You need to expand it to a normal equation. Then you solve that for zeros to get your critical numbers.

Yeah, don't know how to find roots from there lol.

Did you try the product rule :

(uv)' = u'v + uv'

??

Use the rational root theorem. Notice 3/1 is one of the possibilities and use long division to confirm. Then you'll be down to a quadratic 7x^2 -33x +36. Then you can factor the quadratic by using a tree.