Question Number 188822 by sciencestudentW last updated on 07/Mar/23
$${is}\:{it}\:{a}\:{polynomial}? \\ $$$${p}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}^{{x}} −\mathrm{10} \\ $$
Commented by mr W last updated on 07/Mar/23
$${you}\:{seem}\:{not}\:{to}\:{have}\:{read}\:{the} \\ $$$${definition}\:{of}\:{polynomial}. \\ $$
Commented by mr W last updated on 07/Mar/23
Commented by sciencestudentW last updated on 07/Mar/23
$${thanks}\:{sir} \\ $$
Answered by a.lgnaoui last updated on 07/Mar/23
![No; p(x) is in format: (ax^n +b)+x^x (a=n=2) first terme 2x^2 +4x−10 is polynominal but the second terme is not wuih aplication of addition polynomes: Sum,produit [of 2 or more polynomes] is a polynome so; Expression like yours show that is not polynome because the (exposant of x^x ) must be terme ∈Q) x^x =1.x^x (x non defini) exposant x:must be: =m(m∈Z) or= ((p/q))(with p,q∈Z) ax^n : name:Monome ax^n +bx^m or(ax^n +p): binome as (binome of Newton). ax^n +bx^m +cx^p +d(poynome (nb termes>=3)(n ,m ,p∈Z)](https://www.tinkutara.com/question/Q188830.png)
$${No};\:\:{p}\left({x}\right)\:{is}\:{in}\:{format}: \\ $$$$\left({ax}^{{n}} +{b}\right)+{x}^{{x}} \:\:\:\:\left({a}={n}=\mathrm{2}\right) \\ $$$${first}\:{terme}\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{10}\:{is}\: \\ $$$${polynominal}\:\:{but}\:{the}\:{second}\:{terme} \\ $$$${is}\:{not}\: \\ $$$${wuih}\:{aplication}\:{of}\:{addition} \\ $$$${polynomes}: \\ $$$${Sum},{produit}\:\left[{of}\:\mathrm{2}\:{or}\:{more}\:{polynomes}\right]\:{is} \\ $$$${a}\:{polynome} \\ $$$${so};\:\:{Expression}\:{like}\:{yours} \\ $$$${show}\:{that}\:{is}\:{not}\:{polynome} \\ $$$${because}\:{the}\:\left({exposant}\:{of}\:{x}^{{x}} \right) \\ $$$$\left.{must}\:{be}\:\:{terme}\:\in{Q}\right) \\ $$$${x}^{{x}} =\mathrm{1}.{x}^{{x}} \left({x}\:{non}\:{defini}\right) \\ $$$${exposant}\:{x}:{must}\:{be}: \\ $$$$={m}\left({m}\in\mathbb{Z}\right)\:\:\:{or}=\:\left(\frac{{p}}{{q}}\right)\left({with}\:{p},{q}\in\mathbb{Z}\right) \\ $$$${ax}^{{n}} :\:\:\:\:{name}:{Monome} \\ $$$${ax}^{{n}} +{bx}^{{m}} {or}\left({ax}^{{n}} +{p}\right):\:{binome} \\ $$$${as}\:\left({binome}\:{of}\:{Newton}\right). \\ $$$${ax}^{{n}} +{bx}^{{m}} +{cx}^{{p}} +{d}\left({poynome}\right. \\ $$$$\left({nb}\:\:{termes}>=\mathrm{3}\right)\left({n}\:,{m}\:,{p}\in\mathbb{Z}\right) \\ $$
Commented by sciencestudentW last updated on 07/Mar/23
$${that}'{s}\:{great}\:{thanks}\:{sir} \\ $$$$ \\ $$
Commented by mr W last updated on 07/Mar/23
$${wrong}! \\ $$$${x}^{\frac{{p}}{{q}}} \:{is}\:{not}\:{polynomial}\:{for}\:{p},{q}\:\in{Z} \\ $$$$\mathrm{2}{x}^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}\:{is}\:{not}\:{polynomial}. \\ $$
Commented by a.lgnaoui last updated on 07/Mar/23
$${but}\:{in}\:{some}\:{expression}\:{they} \\ $$$${considered}\left[{them}\:{as}\:{polynome}\right. \\ $$$${so}\:\:\:{polynome}\left[{is}\:{only}\:{for}\:\right. \\ $$$${exposants}\:{with}\:{termes}\left[{digits}\left(\mathrm{0},+\mathrm{1},+\mathrm{2},+\mathrm{3},…..\right)\:\right. \\ $$$${thanks}\:{for}\:{distinction}. \\ $$
Commented by mr W last updated on 08/Mar/23
$${yes},\:{only}\:{positive}\:{integer}\:{power}\:{of} \\ $$$${variables}. \\ $$