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Question Number 65289 by naka3546 last updated on 27/Jul/19

Commented by Prithwish sen last updated on 27/Jul/19

$$\mathrm{let}$$$$\mathbf{\Sigma }\mathbf{pqr}=\mathbf{A},\mathbf{\Sigma }\mathbf{pq}=\mathbf{B},\mathbf{\Sigma }\mathbf{p}=\mathbf{C}$$$$\mathrm{Then}\mathrm{from}\mathrm{the}\mathrm{last}\mathrm{three}\mathrm{equation}\mathrm{we}\mathrm{get}$$$$\mathrm{A}+\mathrm{B}+\mathrm{C}=5$$$$\mathrm{A}+2\mathrm{B}+4\mathrm{C}=-3$$$$\mathrm{A}+3\mathrm{B}+9\mathrm{C}=-21$$$$\mathrm{solving}\mathrm{this}\mathrm{we}\mathrm{get}$$$$\mathrm{A}=3,\mathrm{B}=7,\mathrm{C}=-5$$$$\therefore (\mathbf{p}+4)(\mathbf{q}+4)(\mathbf{r}+4)(\mathbf{s}+4)=4\mathbf{A}+16\mathbf{B}+64\mathbf{C}+254$$$$=12+112-320+254=58$$$$\mathrm{please}\mathrm{check}.$$

Answered by mr W last updated on 28/Jul/19

$$pqrs+\mathrm{\Sigma }pqr+\mathrm{\Sigma }pq+\mathrm{\Sigma }p+1=4$$$$\Rightarrow \mathrm{\Sigma }pqr+\mathrm{\Sigma }pq+\mathrm{\Sigma }p=5...\left(i\right)$$$$pqrs+2\mathrm{\Sigma }pqr+4\mathrm{\Sigma }pq+8\mathrm{\Sigma }p+16=8$$$$\Rightarrow \mathrm{\Sigma }pqr+2\mathrm{\Sigma }pq+4\mathrm{\Sigma }p=-3...\left(ii\right)$$$$pqrs+3\mathrm{\Sigma }pqr+9\mathrm{\Sigma }pq+27\mathrm{\Sigma }p+81=16$$$$\Rightarrow \mathrm{\Sigma }pqr+3\mathrm{\Sigma }pq+9\mathrm{\Sigma }p=-21...\left(iii\right)$$$$\Rightarrow \mathrm{\Sigma }pqr=3$$$$\Rightarrow \mathrm{\Sigma }pq=7$$$$\Rightarrow \mathrm{\Sigma }p=-5$$$$$$$$(p+4)(q+4)(r+4)(s+4)$$$$=pqrs+4\mathrm{\Sigma }pqr+16\mathrm{\Sigma }pq+64\mathrm{\Sigma }p+256$$$$=-2+4\times 3+16\times 7+64\times (-5)+256$$$$=-2+12+112-320+256$$$$=58$$$$$$$$generally$$$$(p+n)(q+n)(r+n)(s+n)$$$$=pqrs+n\mathrm{\Sigma }pqr+n^2\mathrm{\Sigma }pq+n^3\mathrm{\Sigma }p+n^4$$$$=-2+3n+7n^2-5n^3+n^4$$$$$$$$==============$$$$p,q,r,sarerootsofeqn.$$$$z^4+5z^3+7z^2-3z-2=0$$

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