# Finding the root of $x^3+3x^2+4$ knowing the roots of $x^3+3x^2$

How can I get the root of $$f(x) = x^3+3x^2+4$$ using the fact that $$-3$$ is a root of $$h(x) = x^3+3x^2$$ (also knowing that $$h$$ is a vertical translation of $$4$$ units of $$f$$)? Does not seems to be hard, but I'm stuck. Already tried using the derivative but could not achieve the result.

• Welcome; Did you draw both graphs? Commented Nov 4, 2022 at 18:38
• notice that if this was possible in general, we would be able to iterate the process and find roots of any polynomial by only knowing the root of $p(x)=x$.
– Alan
Commented Nov 4, 2022 at 18:47
• In general, you can't really transform the roots of $p(x)$ to get roots of $p(x)+a$. This is in contrast to horizontal shitfs: it is a trivial matter to transform the roots of $p(x)$ to get the roots of $p(x+a)$. Commented Nov 4, 2022 at 19:08

There is no easy way to get from the fact that $$x^3+ 3x^2$$ has a root at $$x=-3$$ to finding roots of $$x^3 + 3x^2 + 4$$
There is a root and it is at $$x = -(\sqrt[3]{3+\sqrt{8}} + \sqrt[3]{3-\sqrt{8}} + 1)$$
• At the beginning of your post :$x^3+3x^2$ Commented Nov 4, 2022 at 19:05
• Thanks Doug! I was simply incorrect - I thought I spotted $x=1$ but must have autocompleted to $x^3 + 3 x^2 -4$ in my brain when I decided that ; ) Commented Nov 5, 2022 at 21:06
Knowing the roots to $$x^3+3x^2$$ offers no help in this case since the equation $$x^3+3x^2+4$$ has no rational roots at all. The only roots it does have can be found using the cubic formula or even just Cardano's method, but that is about it. It cannot really be solved via factorization.