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Question Number 78489 by ~blr237~ last updated on 18/Jan/20

$$\mathrm{let}\mathrm{P}\left(\mathrm{x}\right)={\mathrm{x}}^5-209\mathrm{x}+56$$$$\mathrm{Prove}\mathrm{that}\mathrm{there}\mathrm{exist}\mathrm{two}\mathrm{roots}\mathrm{a},\mathrm{b}\mathrm{such}\mathrm{as}\mathrm{ab}=1$$$$\mathrm{Find}\mathrm{out}\mathrm{their}\mathrm{sum}(\mathrm{a}+\mathrm{b}=?)\mathrm{and}\mathrm{deduce}\mathrm{the}\mathrm{decomposition}\mathrm{of}\mathrm{P}\left(\mathrm{x}\right)\mathrm{in}\mathrm{prime}\mathrm{factors}.$$

Answered by MJS last updated on 18/Jan/20

$$2\mathrm{roots}\mathrm{with}ab=1\Rightarrow \mathrm{we}\mathrm{have}\mathrm{a}\mathrm{square}\mathrm{factor}$$$$(x-a)(x-\frac{1}{a})=x^2-(a+\frac{1}{a})x+1=$$$$[\mathrm{let}a+\frac{1}{a}=A]$$$$=x^2-Ax+1$$$$\Rightarrow$$$$\mathrm{the}\mathrm{other}\mathrm{factor}\mathrm{is}$$$$x^3+\alpha x^2+\beta x+56$$$$$$$$x^5-209x+56=(x^2-Ax+1)(x^3+\alpha x^2+\beta x+56)$$$$x^5-209x+56=x^5+(\alpha -A)x^4-(\alpha A-\beta -1)x^3-(\beta A-\alpha -56)x^2+(\beta -56A)x+56$$$$\Rightarrow$$$$\left(1\right)\alpha -A=0$$$$\left(2\right)\alpha A-\beta -1=0$$$$\left(3\right)\beta A-\alpha -56=0$$$$\left(4\right)\beta -56A+209=0$$$$\Rightarrow$$$$A=4\Rightarrow a=2\pm \sqrt{3}\Rightarrow b=2\mp \sqrt{3}$$$$\alpha =4$$$$\beta =15$$$$\Rightarrow$$$$x^5-209x+56=$$$$=(x-2-\sqrt{3})(x-2+\sqrt{3})(x^3+4x^2+15x+56)$$$$\mathrm{and}\mathrm{we}\mathrm{need}{\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{method}\mathrm{for}\mathrm{the}{2}^{\mathrm{nd}}$$$$\mathrm{factor}$$

Commented by ~blr237~ last updated on 18/Jan/20

$$\mathrm{thanks}\mathrm{sir}\mathrm{for}\mathrm{this}\mathrm{part},\mathrm{but}\mathrm{before}\mathrm{reaching}\mathrm{here}\mathrm{we}\mathrm{ought}\mathrm{to}\mathrm{prove}\mathrm{the}\mathrm{existence}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}$$

Commented by MJS last updated on 18/Jan/20

$$\mathrm{I}\mathrm{have}\mathrm{no}\mathrm{idea}\mathrm{how}\mathrm{to}\mathrm{prove}\mathrm{the}\mathrm{existence}\mathrm{of}$$$$\mathrm{the}\mathrm{roots}\mathrm{without}\mathrm{calculating}\mathrm{them}$$

Commented by mr W last updated on 18/Jan/20

$$f^{\prime }\left(x\right)=5x^4-209=0\Rightarrow x_{1,2}=\mp \sqrt{\sqrt{209/5}}$$$$f\left(x_1\right)>0,f\left(x_2\right)<0$$$$f(\to -\mathrm{\infty })\to -\mathrm{\infty },f(\to +\mathrm{\infty })\to +\mathrm{\infty }$$$$\Rightarrow threerealrootsexist.one<-\sqrt{\sqrt{209/5}}$$$$one>\sqrt{\sqrt{209/5}}andthethirdonebetween.$$

Commented by mr W last updated on 18/Jan/20

Commented by mr W last updated on 18/Jan/20

$$thatf\left(x\right)hasonefactor{x}^2+px+1is$$$$theproofthattworootsexistwhose$$$$productis1.$$

Commented by mr W last updated on 18/Jan/20

$$solutionofMJSsirisabsolutely$$$$correctandnice.$$

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