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Question Number 67819 by ajfour last updated on 31/Aug/19

y=x^5 +ax^4 +bx^3 +cx^2 +dx+e  If we let x=t+h  can we find h in terms of a,b,c,d,e  such that  y=(t+R)(t^2 +pt+q)(t^2 +s)  this means two roots are of  opposite sign, of course its  possible by shifting the curve  along x, but can we find the  shift h ?

$${y}={x}^{\mathrm{5}} +{ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e} \\ $$$${If}\:{we}\:{let}\:{x}={t}+{h} \\ $$$${can}\:{we}\:{find}\:{h}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d},{e} \\ $$$${such}\:{that} \\ $$$${y}=\left({t}+{R}\right)\left({t}^{\mathrm{2}} +{pt}+{q}\right)\left({t}^{\mathrm{2}} +{s}\right) \\ $$$${this}\:{means}\:{two}\:{roots}\:{are}\:{of} \\ $$$${opposite}\:{sign},\:{of}\:{course}\:{its} \\ $$$${possible}\:{by}\:{shifting}\:{the}\:{curve} \\ $$$${along}\:{x},\:{but}\:{can}\:{we}\:{find}\:{the} \\ $$$${shift}\:\boldsymbol{{h}}\:? \\ $$

Commented by TawaTawa last updated on 31/Aug/19

Weldone sir on the evetyday trial. It will be fruitful sir.   God bless you sir

$$\mathrm{Weldone}\:\mathrm{sir}\:\mathrm{on}\:\mathrm{the}\:\mathrm{evetyday}\:\mathrm{trial}.\:\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{fruitful}\:\mathrm{sir}.\: \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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