División de Polinomios con residuo

$$\begin{align}&P(x)=Q(x)(x^4-1)+(3x^3+bx^2+cx-2)\\&\\&P(x)=Q(x)(x^2-1)(x^2+1)+(3x^3+bx^2+cx-2)\\&\\&\frac{P(x)}{x^2+1}=Q_1(x)(x^2-1)+\frac{3x^3+bx^2+cx-2}{x^2+1}=Q_1(x)(x^2-1)+3x+b+\frac{(c-3)x-(b+2)}{x^2+1}\\&\\&\frac{P(x)}{x^2-1}=Q_2(x)(x^2+1)+\frac{3x^3+bx^2+cx-2}{x^2-1}=Q_2(x)(x^2-1)+3x+b+\frac{(c+3)x+(b-2)}{x^2-1}\\&\\&\text{Entonces }\\&(c+3)x+(b-2)\equiv 3[(c-3)x-(b+2)]\\&\\&c+3=3(c-3)\to c=6\\&b-2=-3(b+2)\to b=-1\\&\\&\boxed{c+5b=1}\end{align}$$

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