Solve the equation $x^4-6x^3+13x^2-24x+36=0$ if it has multiple root

Given that they have multiple roots

  $$\begin{gathered} Let f\left(x\right)=x^4-6x^3+13x^2-24x+36 \\ Diff. w.r.to \text{'}x\text{'} \\ {f}^{\text{'}}\left(x\right)=4x^3-8x^2+26x-24 \\ By trial and error method \\ {f}^{\text{'}}\left(3\right)=4{\left(3\right)}^3-18{\left(3\right)}^2+26\left(3\right)-24 \\ =186-186=0 \\ Now f\left(3\right)={\left(3\right)}^4-6{\left(3\right)}^3+13{\left(3\right)}^2-24\left(3\right)+36 \\ =234-234=0 \\ \therefore \end{gathered}$$

x-3 is a factor of $f\left(x\right)$ and $f^{\text{'}}\left(x\right)$ then then 3 is a multiple root of order 2 of f(x) By synthetic division 

$$\begin{gathered} x^2+4=0\Rightarrow x^2=-4\Rightarrow x=\pm 2i \\ The roots of given equation are 3, 3, \pm 2i \end{gathered}$$