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Question Number 123265 by Study last updated on 24/Nov/20

Commented by Study last updated on 24/Nov/20

$$whichisapolynomial?why?$$

Commented by mathmax by abdo last updated on 24/Nov/20

$$3)\mathrm{p}\left(\mathrm{x}\right)=\sqrt{3}\mathrm{is}\mathrm{apolynom}\left(\mathrm{csnstant}\right)$$

Answered by MJS_new last updated on 24/Nov/20

$$1)$$$$x<0:p\left(x\right)=\mathrm{i}\sqrt{-x}\times \mathrm{i}\sqrt{-x}={\mathrm{i}}^2\sqrt{(-x)(-x)}=-x$$$$x⩾0:p\left(x\right)=x$$$$\Rightarrow p\left(x\right)=\mid x\mid$$$$2)$$$$p\left(x\right)=\mid x\mid$$$$3)$$$$p\left(x\right)=\sqrt{3}$$$$\mathrm{a}\mathrm{polynome}\mathrm{of}\mathrm{degree}n\mathrm{usually}\mathrm{is}\mathrm{defined}\mathrm{as}$$$$p\left(x\right)=\underset{j=0}{\sum ^n}{c}_j{x}^j\mathrm{with}{c}_j\in \mathbb{C}$$$$\Rightarrow \mathrm{only}p\left(x\right)=\sqrt{3}\mathrm{is}\mathrm{a}\mathrm{polynome};\mathrm{its}\mathrm{degree}\mathrm{is}0$$

Answered by mathmax by abdo last updated on 24/Nov/20

$$2)\mathrm{p}\left(\mathrm{x}\right)=\sqrt{{\mathrm{x}}^2}=\mid \mathrm{x}\mid \mathrm{is}\mathrm{not}\mathrm{polynomial}$$$$\mathrm{let}\mathrm{suppose}\mathrm{it}{\mathrm{p}}^2\left(\mathrm{x}\right)={\mathrm{x}}^2\Rightarrow \mathrm{degp}=1\Rightarrow \mathrm{p}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b}$$$$\mathrm{p}\left(1\right)=1=\mathrm{a}+\mathrm{b}\mathrm{and}\mathrm{p}(-1)=1=-\mathrm{a}+\mathrm{b}\Rightarrow \{\begin{array}{l}\mathrm{a}+\mathrm{b}=1\\ -\mathrm{a}+\mathrm{b}=1\Rightarrow \end{array}$$$$\mathrm{b}=1\Rightarrow \mathrm{a}=0\Rightarrow \mathrm{p}\left(\mathrm{x}\right)=1?\mathrm{impossible}\mathrm{p}\mathrm{is}\mathrm{not}\mathrm{constant}...!$$

Commented by mathmax by abdo last updated on 24/Nov/20

$$1)\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}\mathrm{with}\mathrm{x}⩾0\mathrm{is}\mathrm{not}\mathrm{polynom}$$

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