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Question Number 141814 by gsk2684 last updated on 12/Jul/21

please help me finding the roots of x^5 +5x^4 +20x^3 +60x^2 +120x+120=0?

$${please}\:{help}\:{me}\:{finding}\:{the}\:{roots}\:{of}\:\: \\ $$$${x}^{\mathrm{5}} +\mathrm{5}{x}^{\mathrm{4}} +\mathrm{20}{x}^{\mathrm{3}} +\mathrm{60}{x}^{\mathrm{2}} +\mathrm{120}{x}+\mathrm{120}=\mathrm{0}? \\ $$

Commented by MJS_new last updated on 23/May/21

you don′t believe me?

$$\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{believe}\:\mathrm{me}? \\ $$

Commented by gsk2684 last updated on 23/May/21

Dear MJS, I have registered newly on this platform. I had belief on you . please help me in solving the problem to find the nature of roots and trace of the curve.

$${Dear}\:{MJS},\:{I}\:{have}\:{registered} \\ $$$$\:{newly}\:{on}\:{this}\:{platform}.\:{I}\:{had}\:{belief}\: \\ $$$${on}\:{you}\:.\:{please}\:{help}\:{me}\:{in}\:{solving} \\ $$$${the}\:{problem}\:{to}\:{find}\:{the}\:{nature}\:{of}\: \\ $$$${roots}\:{and}\:{trace}\:{of}\:{the}\:{curve}. \\ $$

Answered by MJS_new last updated on 23/May/21

f(x)=x^5 +5x^4 +20x^3 +60x^2 +120x+120 f ′(x)=5x^4 +20x^3 +60x^2 +120x+120 f ′′(x)=20x^3 +60x^2 +120x+120 f ′′(x)=0 ⇒ x≈−1.59607 f ′(≈−1.59607)≈32.4474 ⇒ f ′(x)>0 ∀x∈R ⇒ ⇒ f(x) is increasing ∀x∈R ⇒ it has exactly one real solution first we try all factors of the constant: ±{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 30, 40, 60, 120} ⇒ none of these solve the equation there′s no solution formula for polynomes of degree 5 or higher ⇒ we must approximate f(0)=120∧f(x) increasing ⇒ search <0 f(−1)=44 f(−2)=8 f(−3)=−78 f(−2.1)=3.77... f(−2.2)=−.968... f(−2.19)=−.465... f(−2.18)=.0299... ... ⇒ x≈−2.18060712404

$${f}\left({x}\right)={x}^{\mathrm{5}} +\mathrm{5}{x}^{\mathrm{4}} +\mathrm{20}{x}^{\mathrm{3}} +\mathrm{60}{x}^{\mathrm{2}} +\mathrm{120}{x}+\mathrm{120} \\ $$$$ \\ $$$${f}\:'\left({x}\right)=\mathrm{5}{x}^{\mathrm{4}} +\mathrm{20}{x}^{\mathrm{3}} +\mathrm{60}{x}^{\mathrm{2}} +\mathrm{120}{x}+\mathrm{120} \\ $$$${f}\:''\left({x}\right)=\mathrm{20}{x}^{\mathrm{3}} +\mathrm{60}{x}^{\mathrm{2}} +\mathrm{120}{x}+\mathrm{120} \\ $$$${f}\:''\left({x}\right)=\mathrm{0}\:\Rightarrow\:{x}\approx−\mathrm{1}.\mathrm{59607} \\ $$$${f}\:'\left(\approx−\mathrm{1}.\mathrm{59607}\right)\approx\mathrm{32}.\mathrm{4474}\:\Rightarrow\:{f}\:'\left({x}\right)>\mathrm{0}\:\forall{x}\in\mathbb{R}\:\Rightarrow \\ $$$$\Rightarrow\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{increasing}\:\forall{x}\in\mathbb{R}\:\Rightarrow\:\mathrm{it}\:\mathrm{has}\:\mathrm{exactly} \\ $$$$\mathrm{one}\:\mathrm{real}\:\mathrm{solution} \\ $$$$ \\ $$$$\mathrm{first}\:\mathrm{we}\:\mathrm{try}\:\mathrm{all}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{constant}: \\ $$$$\pm\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5},\:\mathrm{6},\:\mathrm{8},\:\mathrm{10},\:\mathrm{12},\:\mathrm{15},\:\mathrm{20},\:\mathrm{30},\:\mathrm{40},\:\mathrm{60},\:\mathrm{120}\right\} \\ $$$$\Rightarrow\:\mathrm{none}\:\mathrm{of}\:\mathrm{these}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{polynomes} \\ $$$$\mathrm{of}\:\mathrm{degree}\:\mathrm{5}\:\mathrm{or}\:\mathrm{higher}\:\Rightarrow\:\mathrm{we}\:\mathrm{must}\:\mathrm{approximate} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{120}\wedge{f}\left({x}\right)\:\mathrm{increasing}\:\Rightarrow\:\mathrm{search}\:<\mathrm{0} \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{44} \\ $$$${f}\left(−\mathrm{2}\right)=\mathrm{8} \\ $$$${f}\left(−\mathrm{3}\right)=−\mathrm{78} \\ $$$${f}\left(−\mathrm{2}.\mathrm{1}\right)=\mathrm{3}.\mathrm{77}... \\ $$$${f}\left(−\mathrm{2}.\mathrm{2}\right)=−.\mathrm{968}... \\ $$$${f}\left(−\mathrm{2}.\mathrm{19}\right)=−.\mathrm{465}... \\ $$$${f}\left(−\mathrm{2}.\mathrm{18}\right)=.\mathrm{0299}... \\ $$$$... \\ $$$$\Rightarrow\:{x}\approx−\mathrm{2}.\mathrm{18060712404} \\ $$

Commented by gsk2684 last updated on 24/May/21

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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