allenabby380 allenabby380
  • 08-04-2020
  • Mathematics
contestada

g(n) =n^3 + 4n^2 -n; Find g(-2 - n)

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ItzBlueish
ItzBlueish ItzBlueish
  • 08-04-2020

Answer:

[tex]g=\frac{n^3+4n^2-n}{-2-n}[/tex]; [tex]n\ne \:-2[/tex]

Step-by-step explanation:

[tex]g\left(-2-n\right)=n^3+4n^2-n[/tex]

[tex]\frac{g\left(-2-n\right)}{-2-n}=\frac{n^3}{-2-n}+\frac{4n^2}{-2-n}-\frac{n}{-2-n}[/tex];  [tex]\:n\ne \:-2[/tex]

[tex]g=\frac{n^3+4n^2-n}{-2-n};\quad \:n\ne \:-2[/tex]

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