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Question Number 104181 by mathocean1 last updated on 19/Jul/20

Given P(x)=x^4 +2x^3 −41x^2 +42x+360 Determinate Q(x) a quadratic poly− nom such that: P(x)=(Q(x))^2 −42(Q(x))+360

$${Given} \\ $$$${P}\left({x}\right)={x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{42}{x}+\mathrm{360} \\ $$$${Determinate}\:{Q}\left({x}\right)\:{a}\:{quadratic}\:{poly}− \\ $$$${nom}\:{such}\:{that}: \\ $$$${P}\left({x}\right)=\left({Q}\left({x}\right)\right)^{\mathrm{2}} −\mathrm{42}\left({Q}\left({x}\right)\right)+\mathrm{360} \\ $$

Answered by mr W last updated on 19/Jul/20

let Q(x)=x^2 +bx+c P(x)=(Q(x))^2 −42(Q(x))+360 =x^4 +2bx^3 +(b^2 +2c)x^2 +2bcx+c^2 −42x^2 −42x−42c+360 ⇒c^2 −42c+360=360 ⇒c=42 ⇒2bc−42=42 ⇒bc=42 ⇒b=1 ⇒b^2 +2c−42=41=41 ok! ⇒2b=2=2 ok! ⇒Q(x)=x^2 +x+42

$${let}\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{bx}+{c} \\ $$$${P}\left({x}\right)=\left({Q}\left({x}\right)\right)^{\mathrm{2}} −\mathrm{42}\left({Q}\left({x}\right)\right)+\mathrm{360} \\ $$$$={x}^{\mathrm{4}} +\mathrm{2}{bx}^{\mathrm{3}} +\left({b}^{\mathrm{2}} +\mathrm{2}{c}\right){x}^{\mathrm{2}} +\mathrm{2}{bcx}+{c}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{42}{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:−\mathrm{42}{x}−\mathrm{42}{c}+\mathrm{360} \\ $$$$\Rightarrow{c}^{\mathrm{2}} −\mathrm{42}{c}+\mathrm{360}=\mathrm{360}\:\Rightarrow{c}=\mathrm{42} \\ $$$$\Rightarrow\mathrm{2}{bc}−\mathrm{42}=\mathrm{42}\:\Rightarrow{bc}=\mathrm{42}\:\Rightarrow{b}=\mathrm{1} \\ $$$$\Rightarrow{b}^{\mathrm{2}} +\mathrm{2}{c}−\mathrm{42}=\mathrm{41}=\mathrm{41}\:{ok}! \\ $$$$\Rightarrow\mathrm{2}{b}=\mathrm{2}=\mathrm{2}\:{ok}! \\ $$$$ \\ $$$$\Rightarrow{Q}\left({x}\right)={x}^{\mathrm{2}} +{x}+\mathrm{42} \\ $$

Commented by mathocean1 last updated on 19/Jul/20

thank you sir!

$${thank}\:{you}\:{sir}! \\ $$

Commented by mathocean1 last updated on 19/Jul/20

sir can we also admit that the general form for quadratic polynoms is Q(x)=ax^2 +bx+c ?

$${sir}\:{can}\:{we}\:{also}\:{admit}\:{that}\:{the} \\ $$$${general}\:{form}\:{for}\:{quadratic} \\ $$$${polynoms}\:{is}\:{Q}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:? \\ $$

Commented by mr W last updated on 19/Jul/20

yes, but here it′s obvious that a=1.

$${yes},\:{but}\:{here}\:{it}'{s}\:{obvious}\:{that}\:{a}=\mathrm{1}. \\ $$

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