Question Number 184706 by Ml last updated on 10/Jan/23
![deg[3p(x)+Q(x)]=6 deg[p(x)+x^4 ]=5 deg[(((x^4 +1)p(x^2 ))/(x^3 ∙Q(x)))]=? deg=degree](https://www.tinkutara.com/question/Q184706.png)
$$\mathrm{deg}\left[\mathrm{3p}\left(\mathrm{x}\right)+\mathrm{Q}\left(\mathrm{x}\right)\right]=\mathrm{6} \\ $$$$\mathrm{deg}\left[\mathrm{p}\left(\mathrm{x}\right)+\mathrm{x}^{\mathrm{4}} \right]=\mathrm{5} \\ $$$$\mathrm{deg}\left[\frac{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} \centerdot\mathrm{Q}\left(\mathrm{x}\right)}\right]=? \\ $$$$\mathrm{deg}=\mathrm{degree}\: \\ $$
Answered by Rasheed.Sindhi last updated on 11/Jan/23
![deg[3p(x)+Q(x)]=6;deg[p(x)+x^4 ]=5 deg[(((x^4 +1)p(x^2 ))/(x^3 ∙Q(x)))]=? deg[p(x)+x^4 ]=5⇒deg[p(x)]=5 deg[3p(x)+Q(x)]=6⇒deg[Q(x)]=6 p(x^2 )=10⇒deg[(x^4 +1)p(x^2 )]=14 deg[x^3 ∙Q(x)]=9 deg[(((x^4 +1)p(x^2 ))/(x^3 ∙Q(x)))]=14−9=5 provided that x^3 ∙Q(x) ∣ (x^4 +1)p(x^2 )](https://www.tinkutara.com/question/Q184708.png)
$$\mathrm{deg}\left[\mathrm{3p}\left(\mathrm{x}\right)+\mathrm{Q}\left(\mathrm{x}\right)\right]=\mathrm{6};\mathrm{deg}\left[\mathrm{p}\left(\mathrm{x}\right)+\mathrm{x}^{\mathrm{4}} \right]=\mathrm{5} \\ $$$$\mathrm{deg}\left[\frac{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} \centerdot\mathrm{Q}\left(\mathrm{x}\right)}\right]=? \\ $$$$ \\ $$$$\mathrm{deg}\left[\mathrm{p}\left(\mathrm{x}\right)+\mathrm{x}^{\mathrm{4}} \right]=\mathrm{5}\Rightarrow\mathrm{deg}\left[\mathrm{p}\left(\mathrm{x}\right)\right]=\mathrm{5} \\ $$$$\mathrm{deg}\left[\mathrm{3p}\left(\mathrm{x}\right)+\mathrm{Q}\left(\mathrm{x}\right)\right]=\mathrm{6}\Rightarrow\mathrm{deg}\left[\mathrm{Q}\left(\mathrm{x}\right)\right]=\mathrm{6} \\ $$$$\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} \right)=\mathrm{10}\Rightarrow\mathrm{deg}\left[\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} \right)\right]=\mathrm{14} \\ $$$$\mathrm{deg}\left[\mathrm{x}^{\mathrm{3}} \centerdot\mathrm{Q}\left(\mathrm{x}\right)\right]=\mathrm{9} \\ $$$$\mathrm{deg}\left[\frac{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} \centerdot\mathrm{Q}\left(\mathrm{x}\right)}\right]=\mathrm{14}−\mathrm{9}=\mathrm{5}\: \\ $$$$\mathrm{provided}\:\mathrm{that}\:\mathrm{x}^{\mathrm{3}} \centerdot\mathrm{Q}\left(\mathrm{x}\right)\:\mid\:\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$