Question

Factor and simplify:
(a^2+2a)^2-2(a^2+2a)-3

Answer 1

\[\left(a^2+2a\right)^2:\quad a^4+4a^3+4a^2\]
\[=a^4+4a^3+4a^2-2\left(a^2+2a\right)-3\]
\[\mathrm{Expand}\:-2\left(a^2+2a\right):\quad -2a^2-4a\]
\[=a^4+4a^3+4a^2-2a^2-4a-3\]
\[\mathrm{Add\:similar\:elements:}\:4a^2-2a^2=2a^2\]
\[=a^4+4a^3+2a^2-4a-3\]

Answer 2

Thank you!

Answer 3

\[\mathrm{Factor}\:a^4+4a^3+2a^2-4a-3:\quad \left(a+1\right)^2\left(a+3\right)\left(a-1\right)\]

Answer 4

Is that the answer?

Answer 5

Could you tell me how to factor it?

Answer 6

\[\mathrm{Factor}\:a^4+4a^3+2a^2-4a-3:\quad \left(a+1\right)^2\left(a+3\right)\left(a-1\right)\]
\[a^4+4a^3+2a^2-4a-3\]
Use the rational root theorem
\[a=3,\:\quad a_n=1\]
\[\mathrm{The\:dividers\:of\:}a:\quad 1,\:3,\:\quad
\mathrm{The\:dividers\:of\:}a_n:\quad 1\]
\[\mathrm{Therefore,\:\check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3}{1}\]
\[-\frac{ 1 }{ 1 }\] is a root of the expression,so factor out a+1
\[\mathrm{Compute\:}\frac{a^4+4a^3+2a^2-4a-3}{a+1}\mathrm{\:\to\:get\:the\:rest\:of\:the\:eqution:\quad }a^3+3a^2-a-3\]
\[=\left(a+1\right)\left(a^3+3a^2-a-3\right)\]
\[\mathrm{Factor}\:a^3+3a^2-a-3:\quad \left(a+3\right)\left(a+1\right)\left(a-1\right)\]
\[=\left(a+1\right)\left(a+3\right)\left(a+1\right)\left(a-1\right)\]
\[=\left(a+1\right)^2\left(a+3\right)\left(a-1\right)\]