Respuesta :

[tex]\bf \begin{cases} x=-1\implies &x+1=0\\ x=0\implies &x=0 \end{cases} \\\\\\ \stackrel{\textit{multiplicity of 3}}{(x+1)^3}~~\stackrel{\textit{multiplicity of 1}}{(x)^1}=\stackrel{original~polynomial}{0} \\\\\\ (x^3+3x^2+3x+1)(x)=\stackrel{y}{0}\implies x^4+3x^3+3x^2+x=y[/tex]

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