Question

Factor this

Answer 1

\[a^{2n+1}+b^{2n+1}+a^{2n}b^{2n}+ab\]

Answer 2

@jdoe0001 @jim_thompson5910 @Loser66

Answer 3

hmmm

Answer 4

\(a^{2n+1}+b^{2n+1}+a^{2n}b^{2n}+ab\qquad
\begin{cases}
a^{2n+1}=a^{2n}\cdot a^1\implies a^{2n}a\\
b^{2n+1}=b^{2n}\cdot b^1\implies b^{2n}b
\end{cases}\qquad thus
\\ \quad \\
a^{2n}a+b^{2n}b+a^{2n}b^{2n}+ab
\\ \quad \\
(a^{2n}a+ab)+(b^{2n}b+a^{2n}b^{2n})\impliedby \textit{any common factors?}\)

Answer 5

when you have a^2n(a), that wouldn't be a^4n?

Answer 6

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?

Answer 7

\(\large {
a^n\cdot a^m\implies a^{n+m}
\\ \quad \\
a^{2n+1}=a^{2n}\cdot a^1\implies a^{2n}a
}\)

Answer 8

oh well, then your final answer should be

(a^2n+b)(a+b^2n)?

Answer 9

yeap

Answer 10

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

Answer 11

well... \(\large \begin{cases}
a^{2n+1}=a^{2n}\cdot a^1\implies a^{2n}\cdot a\implies a^{2n}a\\
b^{2n+1}=b^{2n}\cdot b^1\implies b^{2n}\cdot b\implies b^{2n}b
\end{cases}\)

Answer 12

genius thank you