Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Commented by Study last updated on 24/Nov/20

which is a polynomial? why?

$${which}\:{is}\:{a}\:{polynomial}?\:{why}? \\ $$

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

3)p(x)=(√3)is apolynom(csnstant)

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

Answered by MJS_new last updated on 24/Nov/20

1) x<0: p(x)=i(√(−x))×i(√(−x))=i^2 (√((−x)(−x)))=−x x≥0: p(x)=x ⇒ p(x)=∣x∣ 2) p(x)=∣x∣ 3) p(x)=(√3) a polynome of degree n usually is defined as p(x)=Σ_(j=0) ^n c_j x^j with c_j ∈C ⇒ only p(x)=(√3) is a polynome; its degree is 0

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

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

2)p(x)=(√x^2 )=∣x∣ is not polynomial let suppose it p^2 (x)=x^2 ⇒degp=1 ⇒p(x)=ax+b p(1)=1=a+b and p(−1)=1 =−a+b ⇒ { ((a+b=1)),((−a+b=1 ⇒)) :} b=1 ⇒a=0 ⇒p(x)=1 ?impossible p is not constant...!

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

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

1) p(x)=x with x≥0 is not polynom

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com