dtan5457
  • dtan5457
Factor this
Mathematics
schrodinger
  • schrodinger
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dtan5457
  • dtan5457
\[a^{2n+1}+b^{2n+1}+a^{2n}b^{2n}+ab\]
dtan5457
  • dtan5457
jdoe0001
  • jdoe0001
hmmm

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jdoe0001
  • jdoe0001
\(a^{2n+1}+b^{2n+1}+a^{2n}b^{2n}+ab\qquad \begin{cases} a^{2n+1}=a^{2n}\cdot a^1\implies a^{2n}a\\ b^{2n+1}=b^{2n}\cdot b^1\implies b^{2n}b \end{cases}\qquad thus \\ \quad \\ a^{2n}a+b^{2n}b+a^{2n}b^{2n}+ab \\ \quad \\ (a^{2n}a+ab)+(b^{2n}b+a^{2n}b^{2n})\impliedby \textit{any common factors?}\)
dtan5457
  • dtan5457
when you have a^2n(a), that wouldn't be a^4n?
jdoe0001
  • jdoe0001
anyhow \(a^{2n+1}+b^{2n+1}+a^{2n}b^{2n}+ab\qquad \begin{cases} a^{2n+1}=a^{2n}\cdot a^1\implies a^{2n}a\\ b^{2n+1}=b^{2n}\cdot b^1\implies b^{2n}b \end{cases}\qquad thus \\ \quad \\ a^{2n}a+b^{2n}b+a^{2n}b^{2n}+ab \\ \quad \\ (a^{2n}a+ab)+(b^{2n}b+a^{2n}b^{2n}) \\ \quad \\ a({\color{brown}{ a^{2n}+b}})+b^{2n}( b+a^{2n})\implies a({\color{brown}{ a^{2n}+b}})+b^{2n}({\color{brown}{ a^{2n}+b }} )\) see any more common factors?
jdoe0001
  • jdoe0001
\(\large { a^n\cdot a^m\implies a^{n+m} \\ \quad \\ a^{2n+1}=a^{2n}\cdot a^1\implies a^{2n}a }\)
dtan5457
  • dtan5457
oh well, then your final answer should be (a^2n+b)(a+b^2n)?
jdoe0001
  • jdoe0001
yeap
dtan5457
  • dtan5457
that a^2n(a) was confusing for me cause i thought it meant a^2(2n)=a^4n, im just reading it wrong i guess
jdoe0001
  • jdoe0001
well... \(\large \begin{cases} a^{2n+1}=a^{2n}\cdot a^1\implies a^{2n}\cdot a\implies a^{2n}a\\ b^{2n+1}=b^{2n}\cdot b^1\implies b^{2n}\cdot b\implies b^{2n}b \end{cases}\)
dtan5457
  • dtan5457
genius thank you

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