Question

Factor completely 3n^3 + 4n^2 + 24n + 32

Answer 1

In real numbers:

$3n^3+4n^2+24n+32=n^2(3n+4)+8(3n+4)\\\\=(3n+4)(n^2+8)$

In complex numbers:

$=(3n+4)(n^2-(-8))=(3n+4)(n^2-(i\sqrt8)^2)\\\\=(3n+4)(n-i\sqrt8)(n+i\sqrt8)\\\\\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2\\\\=(3n+4)(n-2i\sqrt2)(n+2i\sqrt2)$

Used $(a-b)(a+b)=a^2-b^2$