Question-172632

Question Number 172632 by Mikenice last updated on 29/Jun/22

Commented by peter frank last updated on 29/Jun/22

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mr W last updated on 29/Jun/22

((x^5 +8)/(x+1)) =x(x^5 +8) =x((1−x)^2 x+8) =x((x^2 −2x+1)x+8) =x((2−3x)x+8) =x(2x−3x^2 +8) =x(2x−3(1−x)+8) =5x(x+1) =5

$$\frac{{x}^{\mathrm{5}} +\mathrm{8}}{{x}+\mathrm{1}} \\ $$$$={x}\left({x}^{\mathrm{5}} +\mathrm{8}\right) \\ $$$$={x}\left(\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {x}+\mathrm{8}\right) \\ $$$$={x}\left(\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right){x}+\mathrm{8}\right) \\ $$$$={x}\left(\left(\mathrm{2}−\mathrm{3}{x}\right){x}+\mathrm{8}\right) \\ $$$$={x}\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{8}\right) \\ $$$$={x}\left(\mathrm{2}{x}−\mathrm{3}\left(\mathrm{1}−{x}\right)+\mathrm{8}\right) \\ $$$$=\mathrm{5}{x}\left({x}+\mathrm{1}\right) \\ $$$$=\mathrm{5} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Jun/22

$$\mathcal{N}{ice}\:\boldsymbol{{sir}}! \\ $$

Commented by JDamian last updated on 29/Jun/22

$$={x}\left({x}^{\mathrm{5}} +\mathrm{8}\right)\:\:\:\:\:{shouldn}'{t}\:{be}\:{a}\:\mathrm{3}? \\ $$

Commented by mr W last updated on 29/Jun/22

why? i did: x^2 +x=1 ⇒x(x+1)=1 ⇒(1/(x+1))=x ((x^5 +8)/(x+1))=x(x^5 +8)

$${why}? \\ $$$${i}\:{did}: \\ $$$${x}^{\mathrm{2}} +{x}=\mathrm{1}\:\Rightarrow{x}\left({x}+\mathrm{1}\right)=\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{{x}+\mathrm{1}}={x} \\ $$$$\frac{{x}^{\mathrm{5}} +\mathrm{8}}{{x}+\mathrm{1}}={x}\left({x}^{\mathrm{5}} +\mathrm{8}\right) \\ $$

Commented by Tawa11 last updated on 30/Jun/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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Question-172632

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