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P-x-3-x-2-x-5-x-P-1-




Question Number 172842 by cortano1 last updated on 02/Jul/22
       P(x^3 −x^2 ) = x^(5 )  + x         P(−1)=?
$$\:\:\:\:\:\:\:{P}\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} \right)\:=\:{x}^{\mathrm{5}\:} \:+\:{x}\: \\ $$$$\:\:\:\:\:\:{P}\left(−\mathrm{1}\right)=? \\ $$
Commented by greougoury555 last updated on 02/Jul/22
 ⇒x^3 −x^2 =−1 ; x^3 =x^2 −1  ⇒x^4  = x^3 −x = x^2 −x−1  ⇒x^5  = x^3 −x^2 −x  ⇒x^5 +x = x^2 −1−x^2  =−1    ∴ P(−1)= −1
$$\:\Rightarrow{x}^{\mathrm{3}} −{x}^{\mathrm{2}} =−\mathrm{1}\:;\:{x}^{\mathrm{3}} ={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{4}} \:=\:{x}^{\mathrm{3}} −{x}\:=\:{x}^{\mathrm{2}} −{x}−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{5}} \:=\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x} \\ $$$$\Rightarrow{x}^{\mathrm{5}} +{x}\:=\:{x}^{\mathrm{2}} −\mathrm{1}−{x}^{\mathrm{2}} \:=−\mathrm{1} \\ $$$$\:\:\therefore\:{P}\left(−\mathrm{1}\right)=\:−\mathrm{1} \\ $$

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