[A] $5,0$ e $5,5$.
[B] $4,0$ e $4,5$.
[C] $4,5$ e $5,0$.
[D] $5,5$ e $6,0$.
Solução
Calculando $P(-1)$: $$P(x)=4x^{3}+8x^{2}+x+1$$ $$P(-1)=4 \cdot (-1)^{3}+8 \cdot (-1)^{2}+(-1)+1$$ $$P(-1)=4 \cdot (-1)+8 \cdot 1 -1+1$$ $$P(-1)=-4+8 -1+1$$ $$P(-1)=4$$ Calculando $P(-\frac{1}{3})$: $$P(x)=4x^{3}+8x^{2}+x+1$$ $$P\left(-\frac{1}{3}\right)=4\left(-\frac{1}{3}\right)^{3}+8\left(-\frac{1}{3}\right)^{2}+\left(-\frac{1}{3}\right)+1$$ $$P\left(-\frac{1}{3}\right)=4\left(-\frac{1}{27}\right)+8\left(\frac{1}{9}\right)+\left(-\frac{1}{3}\right)+1$$ $$P\left(-\frac{1}{3}\right)=\left(-\frac{4}{27}\right)+\left(\frac{8}{9}\right)+\left(-\frac{1}{3}\right)+1$$ $$P\left(-\frac{1}{3}\right)=\frac{-4+24-9+27}{27}$$ $$P\left(-\frac{1}{3}\right)=\frac{38}{27}$$ Logo: $$P(-1)+P\left(-\frac{1}{3}\right)=4+\frac{38}{27}=\frac{108+38}{27}=\frac{146}{27}=5,4$$
Resposta: Item A