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Question Number 121585 by Algoritm last updated on 09/Nov/20

Commented by Algoritm last updated on 09/Nov/20

$$\mathbf{answer}:\mathbf{x}=2\sqrt{2}-3$$

Commented by MJS_new last updated on 09/Nov/20

$$x=r{\mathrm{e}}^{\mathrm{i}\theta }$$$$\mathrm{if}\mathrm{we}\mathrm{stay}\mathrm{strictly}\mathrm{real}x\in \mathbb{R}\Rightarrow \theta =0\vee \theta =\pi$$$$\mathrm{usually}\mathrm{we}\mathrm{define}x\in \mathbb{R}:x^{1/3}=\mid x{\mid }^{1/3}\mathrm{sign}\left(x\right)$$$$\mathrm{but}\mathrm{if}{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{in}\mathbb{C}\mathrm{we}\mathrm{use}\mathrm{the}\mathrm{root}$$$$x^{1/3}=r^{1/3}{\mathrm{e}}^{\mathrm{i}\frac{\theta }{3}}\mathrm{which}\mathrm{is}\notin \mathbb{R}\mathrm{for}x\in {\mathbb{R}}^{-}$$$$\mathrm{so}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{decide}.\mathrm{if}\mathrm{we}\mathrm{use}\mathrm{the}{1}^{\mathrm{st}}\mathrm{definition}$$$$\left(\mathrm{strictly}\mathrm{real}\right)\mathrm{we}\mathrm{get}$$$${(x^3+7x^2+16x+5)}^3=(-3x^2-7x+5){(1-2x)}^3$$$$x^9+21x^8+195x^7+1030x^6+3306x^5+6571x^4+7637x^3+4266x^2+1237x+120=0$$$$(x+3)(x^2+6x+1)(x^6+12x^5+68x^4+187x^3+295x^2+159x+40)=0$$$$x_1=-3-2\sqrt{2}$$$$x_2=-3$$$$x_3=-3+2\sqrt{2}$$$$\mathrm{if}\mathrm{we}\mathrm{use}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{definition}\mathrm{we}\mathrm{get}$$$$x_1=-3+2\sqrt{2}$$$$x_{2,3}\approx -3.99799\pm 3.26066\mathrm{i}$$

Commented by Dwaipayan Shikari last updated on 09/Nov/20

$$Sir,howdidyoufactorizeit?$$

Commented by MJS_new last updated on 09/Nov/20

$$-3\mathrm{was}\mathrm{easy}\mathrm{to}\mathrm{find}(\mathrm{try}\mathrm{factors}\mathrm{of}\mathrm{the}$$$$\mathrm{constant}).\mathrm{the}\mathrm{rest}\mathrm{usually}\mathrm{depends}\mathrm{on}\mathrm{good}$$$$\mathrm{luck}.\mathrm{you}\mathrm{can}\mathrm{approximate}\mathrm{the}\mathrm{real}\mathrm{solutions}$$$$\mathrm{and}\mathrm{try}\mathrm{to}\mathrm{see}\mathrm{if}\mathrm{they}‘‘\mathrm{are}{\mathrm{something}}^{″}\mathrm{or}$$$$\mathrm{you}\mathrm{must}\mathrm{use}\mathrm{software}$$$$\mathrm{in}\mathrm{this}\mathrm{case}-3+2\sqrt{2}\mathrm{was}\mathrm{given},\mathrm{so}\mathrm{I}\mathrm{tried}$$$$\mathrm{it}\mathrm{and}\mathrm{then}\mathrm{very}\mathrm{often}\mathrm{the}\mathrm{conjugate}\mathrm{is}\mathrm{also}$$$$\mathrm{a}\mathrm{solution}$$$$\mathrm{for}\mathrm{solutions}\mathrm{that}\mathrm{cannot}\mathrm{be}\mathrm{written}\mathrm{as}$$$$a\pm \sqrt{b}\mathrm{you}\mathrm{have}\mathrm{to}\mathrm{use}\mathrm{software}.\mathrm{or}\mathrm{spend}\mathrm{lots}$$$$\mathrm{of}\mathrm{time}...$$

Commented by Dwaipayan Shikari last updated on 09/Nov/20

$$Supposeagiven{x}^5-x^3+x-2=0$$$$Andweareaskedtofactorizeit,thenwhichmethodweshould$$$$applysir?$$

Commented by MJS_new last updated on 09/Nov/20

$$\mathrm{first},\mathrm{try}\pm 2,\pm 1\Rightarrow \mathrm{no}\mathrm{solution}$$$$\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{real}\mathrm{zeros}.\mathrm{plot}\mathrm{it}\mathrm{or}$$$$\mathrm{go}\mathrm{on}\mathrm{like}\mathrm{this}:$$$$y^{\prime }=5x^4-3x^2+1=0$$$$\mathrm{this}\mathrm{has}\mathrm{no}\mathrm{real}\mathrm{solutions}\Rightarrow y\mathrm{has}\mathrm{no}\mathrm{extremes}$$$$\Rightarrow \mathrm{as}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}{(2n+1)}^{\mathrm{th}}\mathrm{degree}\mathrm{polynomial}\mathrm{it}$$$$\mathrm{must}\mathrm{have}\mathrm{exactly}1\mathrm{real}\mathrm{solution}$$$$\mathrm{if}\mathrm{we}\mathrm{are}\mathrm{lucky}\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}$$$$y=(x^2+ax+b)(x^3+\alpha x^2+\beta x+\gamma )$$$$\mathrm{which}\mathrm{gives}$$$$\left(1\right)a+\alpha =0$$$$\left(2\right)a\alpha +b+\beta +1=0$$$$\left(3\right)a\beta +b\alpha +\gamma =0$$$$\left(4\right)a\gamma +b\beta -1=0$$$$\left(5\right)b\gamma +2=0$$$$\mathrm{solving}\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{easy}\mathrm{but}\mathrm{possible}$$$$\mathrm{we}\mathrm{get}\mathrm{a}‘‘{\mathrm{nice}}^{″}\mathrm{solution}$$$$a=\beta =-1$$$$b=\alpha =1$$$$\gamma =-2$$$$\mathrm{but}\mathrm{in}\mathrm{most}\mathrm{cases}\mathrm{this}{\mathrm{won}}^{\prime }\mathrm{t}\mathrm{lead}\mathrm{to}\mathrm{exact}$$$$\mathrm{values}\mathrm{for}\mathrm{these}\mathrm{factors}.$$$$\mathrm{try}$$$$x^5+2x^3-3x+5$$

Commented by Dwaipayan Shikari last updated on 10/Nov/20

$$Thankingyou$$

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