How many real solutions can an increasing polynomial function have and why ?

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$$\begin{gathered} \mathrm{An} \mathrm{increasing} \mathrm{polynomial} \mathrm{can} \mathrm{have} \mathrm{atmost} \mathrm{one} \mathrm{real} \mathrm{solution} \mathrm{because} \\ \mathrm{value} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \mathrm{keeps} \mathrm{on} \mathrm{incresing} \mathrm{as} \mathrm{we} \mathrm{increase} \mathrm{value} \mathrm{of} \mathrm{x}, \mathrm{hence} \\ \mathrm{it} \mathrm{f}\left(\mathrm{x}\right)=0 \mathrm{for} \mathrm{some} \mathrm{x}=\mathrm{\alpha }, \mathrm{it} \mathrm{cannot} \mathrm{be} 0 \mathrm{again} \mathrm{for} \mathrm{x}>\mathrm{\alpha } \mathrm{because} \mathrm{value} \\ \mathrm{will} \mathrm{increase} \mathrm{from} 0 \mathrm{and} \mathrm{keep} \mathrm{on} \mathrm{increasing}. \end{gathered}$$

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