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Question Number 128712 by Study last updated on 09/Jan/21

if P(x−1)=2x^3 ∙Q(x)+x^2 find ((P(1)−4)/(Q(x)))=??

$${if}\:{P}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{3}} \centerdot{Q}\left({x}\right)+{x}^{\mathrm{2}} \\ $$$${find}\:\frac{{P}\left(\mathrm{1}\right)−\mathrm{4}}{{Q}\left({x}\right)}=?? \\ $$

Commented by Adel last updated on 13/Jan/21

p(1)=p(2−1)=2×2^3 ×Q(x)+2^2 p(1)=16Q(X)+4 P(1)−4=16Q(X) ((P(1)−4)/(Q(X)))=16

$$\mathrm{p}\left(\mathrm{1}\right)=\boldsymbol{\mathrm{p}}\left(\mathrm{2}−\mathrm{1}\right)=\mathrm{2}×\mathrm{2}^{\mathrm{3}} ×\boldsymbol{\mathrm{Q}}\left(\mathrm{x}\right)+\mathrm{2}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{p}}\left(\mathrm{1}\right)=\mathrm{16}\boldsymbol{\mathrm{Q}}\left(\mathrm{X}\right)+\mathrm{4} \\ $$$$\mathrm{P}\left(\mathrm{1}\right)−\mathrm{4}=\mathrm{16Q}\left(\mathrm{X}\right) \\ $$$$\frac{\mathrm{P}\left(\mathrm{1}\right)−\mathrm{4}}{\mathrm{Q}\left(\mathrm{X}\right)}=\mathrm{16} \\ $$

Answered by essafty last updated on 09/Jan/21

16

$$\mathrm{16} \\ $$$$ \\ $$

Commented by Study last updated on 09/Jan/21

what is the practice??

$${what}\:{is}\:{the}\:{practice}?? \\ $$

Answered by hknkrc46 last updated on 09/Jan/21

Commented by Study last updated on 10/Jan/21

how ((16Q(2))/(Q(x)))=16 ??? or ((Q(2))/(Q(x)))=1 ?? why??

$${how}\:\frac{\mathrm{16}{Q}\left(\mathrm{2}\right)}{{Q}\left({x}\right)}=\mathrm{16}\:???\:{or}\:\frac{{Q}\left(\mathrm{2}\right)}{{Q}\left({x}\right)}=\mathrm{1}\:??\: \\ $$$${why}?? \\ $$

Commented by Study last updated on 11/Jan/21

why??

$${why}?? \\ $$

Commented by Study last updated on 11/Jan/21

plz ans me

$${plz}\:{ans}\:{me} \\ $$

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