mrmickle60
mrmickle60 mrmickle60
  • 18-10-2018
  • Mathematics
contestada

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

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gmany
gmany gmany
  • 19-10-2018

In real numbers:

[tex]3n^3+4n^2+24n+32=n^2(3n+4)+8(3n+4)\\\\=(3n+4)(n^2+8)[/tex]

In complex numbers:

[tex]=(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)[/tex]

Used [tex](a-b)(a+b)=a^2-b^2[/tex]

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