Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 78650 by Pratah last updated on 19/Jan/20

Commented by mr W last updated on 19/Jan/20

$$=x^2+1$$

Commented by john santu last updated on 20/Jan/20

$$sorrysir,yesterdayalready$$$$slept$$

Commented by Pratah last updated on 19/Jan/20

$$\mathrm{sir}\mathrm{john}\mathrm{santu}$$

Commented by Pratah last updated on 19/Jan/20

$$\mathrm{please}$$

Answered by mind is power last updated on 19/Jan/20

$$\mathrm{p}={3\mathrm{x}}^4+{2\mathrm{x}}^3+{\mathrm{x}}^2+2\mathrm{x}-2={3\mathrm{x}}^4-3+{2\mathrm{x}}^3+2\mathrm{x}+{\mathrm{x}}^2+1=3({\mathrm{x}}^2-1)({\mathrm{x}}^2+1)$$$$+2\mathrm{x}(1+{\mathrm{x}}^2)+{\mathrm{x}}^2+1=({\mathrm{x}}^2+1)(3({\mathrm{x}}^2-1)+2\mathrm{x}+1)=({\mathrm{x}}^2+1)({3\mathrm{x}}^2+2\mathrm{x}-2)$$$$\mathrm{Q}={\mathrm{x}}^5+{\mathrm{x}}^4-{\mathrm{x}}^3-2\mathrm{x}-1={\mathrm{x}}^5+{\mathrm{x}}^3+{\mathrm{x}}^4-1-2\mathrm{x}-{2\mathrm{x}}^3$$$$={\mathrm{x}}^3(1+{\mathrm{x}}^2)+({\mathrm{x}}^2-1)({\mathrm{x}}^2+1)-2\mathrm{x}(1+{\mathrm{x}}^2)=(1+{\mathrm{x}}^2)({\mathrm{x}}^3+{\mathrm{x}}^2-1-2\mathrm{x})$$$$=(1+{\mathrm{x}}^2)({\mathrm{x}}^3+{\mathrm{x}}^2-2\mathrm{x}-1)$$$$\mathrm{So}(1+{\mathrm{x}}^2)\mid \gcd (\mathrm{p},\mathrm{q})$$$${\mathrm{x}}^3+{\mathrm{x}}^2-2\mathrm{x}-1\mathrm{and}{3\mathrm{x}}^2+2\mathrm{x}-2$$$$\mathrm{has}\mathrm{dufferent}\mathrm{roots}\Rightarrow \mathrm{primes}$$$$\Rightarrow$$$$\gcd =1+{\mathrm{x}}^2$$

Commented by Pratah last updated on 19/Jan/20

$$\mathrm{thanks}$$

Answered by Rasheed.Sindhi last updated on 19/Jan/20

$$\gcd ({\mathrm{x}}^5+{\mathrm{x}}^4-{\mathrm{x}}^3-2\mathrm{x}-1,{3\mathrm{x}}^4+{2\mathrm{x}}^3+{\mathrm{x}}^2+2\mathrm{x}-2)$$$$=\gcd (3({\mathrm{x}}^5+{\mathrm{x}}^4-{\mathrm{x}}^3-2\mathrm{x}-1),{3\mathrm{x}}^4+{2\mathrm{x}}^3+{\mathrm{x}}^2+2\mathrm{x}-2)$$$$\left|\begin{array}{cc}& \mathrm{x}+1\\ {3\mathrm{x}}^4+{2\mathrm{x}}^3+{\mathrm{x}}^2+2\mathrm{x}-2)& +{3\mathrm{x}}^5+{3\mathrm{x}}^4-{3\mathrm{x}}^3+{0\mathrm{x}}^2-6\mathrm{x}-3\\ & \underset{-}{+}{3\mathrm{x}}^5\underset{-}{+}{2\mathrm{x}}^4\underset{-}{+}{\mathrm{x}}^3\underset{-}{+}{2\mathrm{x}}^2\underset{+}{-}2\mathrm{x}\\ & 3({\mathrm{x}}^4-{4\mathrm{x}}^3-{2\mathrm{x}}^2-4\mathrm{x}-3)\\ & ={3\mathrm{x}}^4-{12\mathrm{x}}^3-{6\mathrm{x}}^2-12\mathrm{x}-9\\ & \underset{-}{+}{3\mathrm{x}}^4\underset{-}{+}{2\mathrm{x}}^3\underset{-}{+}{\mathrm{x}}^2\underset{-}{+}2\mathrm{x}\underset{+}{-}2\\ & -{14\mathrm{x}}^3-{7\mathrm{x}}^2-14\mathrm{x}-7\\ & =-7({2\mathrm{x}}^3+{\mathrm{x}}^2+2\mathrm{x}+1)\end{array}\right|$$$$$$$$\left|\begin{array}{cc}& \mathrm{x}+1\\ {6\mathrm{x}}^3+{3\mathrm{x}}^2+6\mathrm{x}+3)& +{6\mathrm{x}}^4+{4\mathrm{x}}^3+{2\mathrm{x}}^2+4\mathrm{x}-4\\ & \underset{-}{+}{6\mathrm{x}}^4\underset{-}{+}{3\mathrm{x}}^3\underset{-}{+}{6\mathrm{x}}^2\underset{-}{+}3\mathrm{x}\\ & {\mathrm{x}}^3-{4\mathrm{x}}^2+\mathrm{x}-4\\ & 6({\mathrm{x}}^3-{4\mathrm{x}}^2+\mathrm{x}-4)\\ & ={6\mathrm{x}}^3-{24\mathrm{x}}^2+6\mathrm{x}-24\\ & \underset{-}{+}{6\mathrm{x}}^3\underset{-}{+}{3\mathrm{x}}^2\underset{-}{+}6\mathrm{x}\underset{-}{+}3\\ & -{27\mathrm{x}}^2-27\\ & =-27({\mathrm{x}}^2+1)\end{array}\right|$$$$$$$$\left|\begin{array}{cc}& 6\mathrm{x}+3\\ {\mathrm{x}}^2+1)& {6\mathrm{x}}^3+{3\mathrm{x}}^2+6\mathrm{x}+3\\ & \underset{-}{+}{6\mathrm{x}}^3\underset{-}{+}6\mathrm{x}\\ & {3\mathrm{x}}^2+3\\ & \underset{-}{+}{3\mathrm{x}}^2\underset{-}{+}3\\ & 0\end{array}\right|$$$$\gcd ={\mathrm{x}}^2+1$$

Commented by Pratah last updated on 19/Jan/20

$$\mathrm{great}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com