factorise fully x^3 + 2x^2 - 5x - 72

If this is true radio you need to tell anon how you know this.  

$${{\mathtt{x}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{72}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{{\mathtt{19}}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{10\,342}}}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{919}}}{{\mathtt{27}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\sqrt{{\mathtt{10\,342}}}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{919}}}{{\mathtt{27}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\left({\frac{{\sqrt{{\mathtt{10\,342}}}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{919}}}{{\mathtt{27}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\times\,}}\left({\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{19}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{10\,342}}}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{919}}}{{\mathtt{27}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\\
{\mathtt{x}} = {\left({\frac{{\sqrt{{\mathtt{10\,342}}}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{919}}}{{\mathtt{27}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{19}}}{\left({\mathtt{9}}{\mathtt{\,\times\,}}{\left({\frac{{\sqrt{{\mathtt{10\,342}}}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{919}}}{{\mathtt{27}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{2.965\: \!540\: \!944\: \!547\: \!592\: \!7}}{\mathtt{\,-\,}}{\mathtt{3.085\: \!633\: \!727\: \!967\: \!809\: \!5}}{i}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{2.965\: \!540\: \!944\: \!547\: \!592\: \!7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3.085\: \!633\: \!727\: \!967\: \!809\: \!5}}{i}\\
{\mathtt{x}} = {\mathtt{3.931\: \!081\: \!889\: \!095\: \!185\: \!4}}\\
\end{array} \right\}$$

Just doing this on the web2 calc, there does appear to be one real  but it is not an integer.That is 

(x-3.931...) is a factor  3.931 is an approximation

Here is another approximation of the factorisation.  It is the same.

http://www.wolframalpha.com/input/?i=factorise+x%5E3%2B2x%5E2-5x-72

This cubic cannot be factorised so this is its simplest form.

There is only one real value for x and 2 imaginary.