Question Number 206037 by Davidtim last updated on 05/Apr/24
$${is}\:{it}\:{a}\:{polynomial}? \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{2}} }+\mathrm{10} \\ $$
Answered by Frix last updated on 05/Apr/24
$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{2}} }+\mathrm{10} \\ $$$${x}\in\mathbb{R}\:\Rightarrow\:\sqrt{{x}^{\mathrm{2}} }=\mid{x}\mid \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{10};\:{x}<\mathrm{0}}\\{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{10};\:{x}\geqslant\mathrm{0}}\end{cases} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{polynomial}. \\ $$
Answered by BaliramKumar last updated on 05/Apr/24
$$\mathrm{yes} \\ $$