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Question Number 161391 by cortano last updated on 17/Dec/21

  f^( 3) (x)+x^2  f(x)=2x^3 +4x^2 +3x+1   ∀x∈R

$$\:\:{f}^{\:\mathrm{3}} \left({x}\right)+{x}^{\mathrm{2}} \:{f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1} \\ $$$$\:\forall{x}\in\mathbb{R}\: \\ $$

Commented by Rasheed.Sindhi last updated on 17/Dec/21

f(x) is not a polynomial.

$${f}\left({x}\right)\:{is}\:{not}\:{a}\:{polynomial}. \\ $$

Commented by mr W last updated on 18/Dec/21

but if the question is just a little bit  different, for example  f^3 (x)+x^2 f(x)=2x^3 +4x^2 +5x+1  then f(x) is no polynomial more.  f(x)=(((√((x^3 +2x^2 +((5x)/2)+(1/2))^2 +(x^6 /(27))))+(x^3 +2x^2 +((5x)/2)+(1/2))))^(1/3)           −(((√((x^3 +2x^2 +((5x)/2)+(1/2))^2 +(x^6 /(27))))−(x^3 +2x^2 +((5x)/2)+(1/2))))^(1/3)

$${but}\:{if}\:{the}\:{question}\:{is}\:{just}\:{a}\:{little}\:{bit} \\ $$$${different},\:{for}\:{example} \\ $$$${f}^{\mathrm{3}} \left({x}\right)+{x}^{\mathrm{2}} {f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1} \\ $$$${then}\:{f}\left({x}\right)\:{is}\:{no}\:{polynomial}\:{more}. \\ $$$${f}\left({x}\right)=\sqrt[{\mathrm{3}}]{\sqrt{\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{x}^{\mathrm{6}} }{\mathrm{27}}}+\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:\:\:\:\:−\sqrt[{\mathrm{3}}]{\sqrt{\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{x}^{\mathrm{6}} }{\mathrm{27}}}−\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{5}{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$

Commented by mr W last updated on 18/Dec/21

yes, it is!   assume it is, then f(x)=ax+b.  comparing the first and last term  it′s easy to see: a=1, b=1.   now check:  f(x)=x+1  (x+1)^3 =   x^3 +3x^2 +3x+1  x^2 (x+1)=x^3 +x^2   −−−−−−−−−−−−−  Σ:              2x^3 +4x^2 +3x+1 ✓

$${yes},\:{it}\:{is}!\: \\ $$$${assume}\:{it}\:{is},\:{then}\:{f}\left({x}\right)={ax}+{b}. \\ $$$${comparing}\:{the}\:{first}\:{and}\:{last}\:{term} \\ $$$${it}'{s}\:{easy}\:{to}\:{see}:\:{a}=\mathrm{1},\:{b}=\mathrm{1}.\: \\ $$$${now}\:{check}: \\ $$$${f}\left({x}\right)={x}+\mathrm{1} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\:\:\:{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)={x}^{\mathrm{3}} +{x}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−− \\ $$$$\Sigma:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\:\checkmark \\ $$

Commented by Rasheed.Sindhi last updated on 18/Dec/21

You′re right sir,I mistook.

$${You}'{re}\:{right}\:{sir},{I}\:{mistook}. \\ $$

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