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let-P-x-x-5-209x-56-Prove-that-there-exist-two-roots-a-b-such-as-ab-1-Find-out-their-sum-a-b-and-deduce-the-decomposition-of-P-x-in-prime-factors-

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Question Number 78489 by ~blr237~ last updated on 18/Jan/20
let P(x)= x^5 −209x+56 Prove that there exist two roots a,b such as ab=1 Find out their sum ( a+b=?) and deduce the decomposition of P(x) in prime factors.
$$\mathrm{let}\:\:\mathrm{P}\left(\mathrm{x}\right)=\:\mathrm{x}^{\mathrm{5}} −\mathrm{209x}+\mathrm{56}\: \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{two}\:\mathrm{roots}\:\:\mathrm{a},\mathrm{b}\:\mathrm{such}\:\mathrm{as}\:\:\:\mathrm{ab}=\mathrm{1} \\ $$$$\mathrm{Find}\:\mathrm{out}\:\mathrm{their}\:\mathrm{sum}\:\left(\:\mathrm{a}+\mathrm{b}=?\right)\:\:\mathrm{and}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{decomposition}\:\mathrm{of}\:\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{in}\:\mathrm{prime}\:\mathrm{factors}. \\ $$
Answered by MJS last updated on 18/Jan/20
2 roots with ab=1 ⇒ we have a square factor (x−a)(x−(1/a))=x^2 −(a+(1/a))x+1= [let a+(1/a)=A] =x^2 −Ax+1 ⇒ the other factor is x^3 +αx^2 +βx+56 x^5 −209x+56=(x^2 −Ax+1)(x^3 +αx^2 +βx+56) x^5 −209x+56=x^5 +(α−A)x^4 −(αA−β−1)x^3 −(βA−α−56)x^2 +(β−56A)x+56 ⇒ (1) α−A=0 (2) αA−β−1=0 (3) βA−α−56=0 (4) β−56A+209=0 ⇒ A=4 ⇒ a=2±(√3) ⇒ b=2∓(√3) α=4 β=15 ⇒ x^5 −209x+56= =(x−2−(√3))(x−2+(√3))(x^3 +4x^2 +15x+56) and we need Cardano′s method for the 2^(nd) factor
$$\mathrm{2}\:\mathrm{roots}\:\mathrm{with}\:{ab}=\mathrm{1}\:\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{square}\:\mathrm{factor} \\ $$$$\left({x}−{a}\right)\left({x}−\frac{\mathrm{1}}{{a}}\right)={x}^{\mathrm{2}} −\left({a}+\frac{\mathrm{1}}{{a}}\right){x}+\mathrm{1}= \\ $$$$\:\:\:\:\:\left[\mathrm{let}\:{a}+\frac{\mathrm{1}}{{a}}={A}\right] \\ $$$$={x}^{\mathrm{2}} −{Ax}+\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{factor}\:\mathrm{is} \\ $$$${x}^{\mathrm{3}} +\alpha{x}^{\mathrm{2}} +\beta{x}+\mathrm{56} \\ $$$$ \\ $$$${x}^{\mathrm{5}} −\mathrm{209}{x}+\mathrm{56}=\left({x}^{\mathrm{2}} −{Ax}+\mathrm{1}\right)\left({x}^{\mathrm{3}} +\alpha{x}^{\mathrm{2}} +\beta{x}+\mathrm{56}\right) \\ $$$${x}^{\mathrm{5}} −\mathrm{209}{x}+\mathrm{56}={x}^{\mathrm{5}} +\left(\alpha−{A}\right){x}^{\mathrm{4}} −\left(\alpha{A}−\beta−\mathrm{1}\right){x}^{\mathrm{3}} −\left(\beta{A}−\alpha−\mathrm{56}\right){x}^{\mathrm{2}} +\left(\beta−\mathrm{56}{A}\right){x}+\mathrm{56} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:\:\alpha−{A}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\alpha{A}−\beta−\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:\beta{A}−\alpha−\mathrm{56}=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:\beta−\mathrm{56}{A}+\mathrm{209}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${A}=\mathrm{4}\:\Rightarrow\:{a}=\mathrm{2}\pm\sqrt{\mathrm{3}}\:\Rightarrow\:{b}=\mathrm{2}\mp\sqrt{\mathrm{3}} \\ $$$$\alpha=\mathrm{4} \\ $$$$\beta=\mathrm{15} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{5}} −\mathrm{209}{x}+\mathrm{56}= \\ $$$$=\left({x}−\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({x}−\mathrm{2}+\sqrt{\mathrm{3}}\right)\left({x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{56}\right) \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{need}\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{method}\:\mathrm{for}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \\ $$$$\mathrm{factor} \\ $$
Commented by ~blr237~ last updated on 18/Jan/20
thanks sir for this part , but before reaching here we ought to prove the existence of a and b
$$\mathrm{thanks}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{this}\:\mathrm{part}\:,\:\mathrm{but}\:\mathrm{before}\:\mathrm{reaching}\:\mathrm{here}\:\mathrm{we}\:\mathrm{ought}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{existence}\:\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b} \\ $$
Commented by MJS last updated on 18/Jan/20
I have no idea how to prove the existence of the roots without calculating them
$$\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{idea}\:\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{existence}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{without}\:\mathrm{calculating}\:\mathrm{them} \\ $$
Commented by mr W last updated on 18/Jan/20
f′(x)=5x^4 −209=0 ⇒x_(1,2) =∓(√(√(209/5))) f(x_1 )>0, f(x_2 )<0 f(→−∞)→−∞, f(→+∞)→+∞ ⇒three real roots exist. one <−(√(√(209/5))) one >(√(√(209/5))) and the third one between.
$${f}'\left({x}\right)=\mathrm{5}{x}^{\mathrm{4}} −\mathrm{209}=\mathrm{0}\:\Rightarrow{x}_{\mathrm{1},\mathrm{2}} =\mp\sqrt{\sqrt{\mathrm{209}/\mathrm{5}}} \\ $$$${f}\left({x}_{\mathrm{1}} \right)>\mathrm{0},\:{f}\left({x}_{\mathrm{2}} \right)<\mathrm{0} \\ $$$${f}\left(\rightarrow−\infty\right)\rightarrow−\infty,\:{f}\left(\rightarrow+\infty\right)\rightarrow+\infty \\ $$$$\Rightarrow{three}\:{real}\:{roots}\:{exist}.\:{one}\:<−\sqrt{\sqrt{\mathrm{209}/\mathrm{5}}} \\ $$$${one}\:>\sqrt{\sqrt{\mathrm{209}/\mathrm{5}}}\:{and}\:{the}\:{third}\:{one}\:{between}. \\ $$
Commented by mr W last updated on 18/Jan/20
Commented by mr W last updated on 18/Jan/20
that f(x) has one factor x^2 +px+1 is the proof that two roots exist whose product is 1.
$${that}\:{f}\left({x}\right)\:{has}\:{one}\:{factor}\:{x}^{\mathrm{2}} +{px}+\mathrm{1}\:{is} \\ $$$${the}\:{proof}\:{that}\:{two}\:{roots}\:{exist}\:{whose} \\ $$$${product}\:{is}\:\mathrm{1}. \\ $$
Commented by mr W last updated on 18/Jan/20
solution of MJS sir is absolutely correct and nice.
$${solution}\:{of}\:{MJS}\:{sir}\:{is}\:{absolutely} \\ $$$${correct}\:{and}\:{nice}. \\ $$

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